NCERT Solutions for Exercise 7.4 Class 12 Maths Chapter 7 - Integrals

In NCERT solutions for Class 12 Maths chapter 7 exercise 7.4, logarithmic functions, parabolic functions etc. are discussed in detail. These concepts are going to help in subsequent Class 12 chapters also. Exercise 7.4 Class 12 Maths is an extension of the earlier exercises with a slightly more difficulty level. NCERT solutions for Class 12 Maths chapter 7 exercise 7.4 provided below are of good quality prepared by subject matter experts. Questions from this NCERT book exercise can be solved by students to increase their speed with accuracy in Integrals. Since speed is also a parameter in scoring well in exams like JEE Main. Also the NCERT chapter Integrals consists of other exercises which are as follows.

  • Integrals Exercise 7.1

  • Integrals Exercise 7.2

  • Integrals Exercise 7.3

  • Integrals Exercise 7.5

  • Integrals Exercise 7.6

  • Integrals Exercise 7.7

  • Integrals Exercise 7.8

  • Integrals Exercise 7.9

  • Integrals Exercise 7.10

  • Integrals Exercise 7.11

  • Integrals Miscellaneous Exercise

Integrals Class 12 Chapter 7 Exercise 7.4

Question:1 Integrate the functions \frac{3x^ 2 }{x^6 + 1 }

Answer:

The given integral can be calculated as follows

Let x^3 = t
, therefore, 3x^2 dx =dt

\Rightarrow \int\frac{3x^2}{x^6+1}=\int \frac{dt}{t^2+1}

\\=\tan^{-1} t +C\\ =tan^{-1}(x^3)+C

Question:2 Integrate the functions \frac{1}{\sqrt { 1+ 4 x^2 }}

Answer:

\frac{1}{\sqrt { 1+ 4 x^2 }}
let suppose 2x = t
therefore 2dx = dt

\int \frac{1}{\sqrt{1+4x^2}} =\frac{1}{2}\int \frac{dt}{1+t^2}
\\=\frac{1}{2}[\log\left | t+\sqrt{1+t^2} \right |]+C\\ =\frac{1}{2}\log\left | 2x+\sqrt{4x^2+1} \right |+C .................using formula \int\frac{1}{\sqrt{x^2+a^2}}dt = \log\left | x+\sqrt{x^2+a^2} \right |

Question:3 Integrate the functions \frac{1}{\sqrt { ( 2- x)^2+ 1 }}

Answer:

\frac{1}{\sqrt { ( 2- x)^2+ 1 }}

let suppose 2-x =t
then, -dx =dt
\Rightarrow \int\frac{1}{\sqrt{(2-x)^2+1}}dx = -\int \frac{1}{\sqrt{t^2+1}}dt

using the identity

\int \frac{1}{\sqrt{x^2+1}}dt=log\left | x+\sqrt{x^2+1} \right |

\\= -\log\left | t+\sqrt{t^2+1} \right |+C\\ =-\log\left | 2-x+\sqrt{(2-x)^2+1} \right |+C\\ =\log \left | \frac{1}{(2-x)+\sqrt{x^2-4x+5}} \right |+C

Question:4 Integrate the functions \frac{1}{\sqrt {9 - 25 x^2 }}

Answer:

\frac{1}{\sqrt {9 - 25 x^2 }}
Let assume 5x =t,
then 5dx = dt

\Rightarrow \int \frac{1}{\sqrt{9-25x^2}}=\frac{1}{5}\int \frac{1}{\sqrt{9-t^2}}dt
\\=\frac{1}{5}\int \frac{1}{\sqrt{3^2-t^2}}dt\\ =\frac{1}{5}\sin^{-1}(\frac{t}{3})+C\\ =\frac{1}{5}\sin^{-1}(\frac{5x}{3})+C

The above result is obtained using the identity

\\\int \frac{1}{\sqrt{a^2-x^2}}dt\\ =\frac{1}{a}sin^{-1}\frac{x}{a}

Question:5 Integrate the functions \frac{3x }{1+ 2 x ^ 4 }

Answer:

\frac{3x }{1+ 2 x ^ 4 }


Let {\sqrt{2}}x^2 = t
\therefore 2\sqrt{2}xdx =dt

The integration can be done as follows

\Rightarrow \int \frac{3x}{1+2x^4}= \frac{3}{2\sqrt{2}}\int \frac{dt}{1+t^2}
\\= \frac{3}{2\sqrt{2}}[\tan^{-1}t]+C\\ =\frac{3}{2\sqrt{2}}[\tan^{-1}(\sqrt{2}x^2)]+C

Question:6 Integrate the functions \frac{x ^ 2 }{1- x ^ 6 }

Answer:

\frac{x ^ 2 }{1- x ^ 6 }

let x^3 =t
then 3x^2dx =dt

using the special identities we can simplify the integral as follows

\int \frac{x^2}{1-x^6}dx =\frac{1}{3}\int \frac{dt}{1-t^2}
=\frac{1}{3}[\frac{1}{2}\log\left | \frac{1+t}{1-t} \right |]+C\\ =\frac{1}{6}\log\left | \frac{1+x^3}{1-x^3} \right |+C

Question:7 Integrate the functions \frac{x-1 }{\sqrt { x^2 -1 }}

Answer:


We can write above eq as
\frac{x-1 }{\sqrt { x^2 -1 }} =\int \frac{x}{\sqrt{x^2-1}}dx-\int \frac{1}{\sqrt{x^2-1}}dx ............................................(i)

for \int \frac{x}{\sqrt{x^2-1}}dx let x^2-1 = t \Rightarrow 2xdx =dt

\therefore \int \frac{x}{\sqrt{x^2-1}}dx=\frac{1}{2}\int \frac{dt}{\sqrt{t}}
\\=\frac{1}{2}\int t^{1/2}dt\\ =\frac{1}{2}[2t^{1/2}]\\ =\sqrt{t}\\ =\sqrt{x^2-1}
Now, by using eq (i)
=\int \frac{x}{\sqrt{x^2-1}}dx-\int \frac{1}{\sqrt{x^2-1}}dx
\\=\sqrt{x^2-1}-\int \frac{1}{\sqrt{x^2}-1}dx\\ =\sqrt{x^2-1}-\log\left | x+\sqrt{x^2-1} \right |+C

Question:8 Integrate the functions \frac{x ^ 2 }{\sqrt { x^6 + a ^ 6 }}

Answer:

The integration can be down as follows

\frac{x ^ 2 }{\sqrt { x^6 + a ^ 6 }}
let x^3 = t \Rightarrow 3x^2dx =dt

\therefore \frac{x^2}{\sqrt{x^6+a^6}}=\frac{1}{3}\int \frac{dt}{\sqrt{t^2+(a^3)^2}}
\\=\frac{1}{3}\log\left | t+\sqrt{t^2+a^6} \right |+C\\ =\frac{1}{3}\log\left | x^3+\sqrt{x^6+a^6} \right |+C ........................using \int \frac{dx}{\sqrt{x^2+a^2}} = \log\left | x+\sqrt{x^2+a^2} \right |

Question:9 Integrate the functions \frac{\sec ^ 2 x }{\sqrt { \tan ^ 2 x+ 4 }}

Answer:

The integral can be evaluated as follows

\frac{\sec ^ 2 x }{\sqrt { \tan ^ 2 x + 4 }}
let \tan x =t \Rightarrow sec^2x dx =dt

\Rightarrow \int \frac{\sec^2x}{\sqrt{\tan^2x+4}}dx = \int \frac{dt}{\sqrt{t^2+2^2}}
\\= \log\left | t+\sqrt{t^2+4} \right |+C\\ =\log \left | \tan x+\sqrt{ tan^2x+4} \right |+C

Question:10 Integrate the functions \frac{1 }{ \sqrt { x ^ 2 + 2 x + 2 }}

Answer:

\frac{1 }{ \sqrt { x ^ 2 + 2 x + 2 }}
the above equation can be also written as,
=\int\frac{1}{\sqrt{(1+x)^2+1^2}}dx
let 1+x = t
then dx = dt
therefore,

\\=\int\frac{1}{\sqrt{t^2+1^2}}dx\\ =\log\left | t+\sqrt{t^2+1} \right |+C\\ =\log\left | (1+x)+\sqrt{(1+x)^2+1} \right |+C\\ =\log\left | (1+x)+\sqrt{(x^2+2x+2} \right |+C

Question:11 Integrate the functions \frac{1}{9 x ^2 + 6x + 5 }

Answer:

\frac{1}{9 x ^2 + 6x + 5 }
this denominator can be written as
9x^2+6x+5=9[x^2+\frac{2}{3}x+\frac{5}{9}]\\=9[(x+\frac{1}{3})^2+(\frac{2}{3})^2] Now,
\frac{1}{9}\int \frac{1}{(x+\frac{1}{3})^2+(\frac{2}{3})^2}dx =\frac{1}{9} [\frac{3}{2}\tan^{-1}(\frac{(x+1/3)}{2/3})] +C\\=\frac{1}{6} \tan^{-1}(\frac{3x+1}{2})] +C
......................................by using the form (\int \frac{1}{x^2+a^2}=\frac{1}{a}\tan^{-1}(\frac{x}{a}))

Question:12 Integrate the functions \frac{1}{\sqrt{ 7-6x - x ^ 2 }}

Answer:

the denominator can be also written as,
7-6x-x^2=16-(x^2+6x+9)
=4^2-(x+3)^2

therefore

\int \frac{1}{\sqrt{7-6x-x^2}}dx=\int \frac{1}{\sqrt{4^2-(x+3)^2}}dx
Let x+3 = t
then dx =dt

\Rightarrow \int \frac{1}{\sqrt{4^2-(x+3)^2}}dx=\int \frac{1}{\sqrt{4^2-t^2}}dt ......................................using formula \int \frac{1}{\sqrt{a^2-x^2}}=\sin^{-1}(\frac{x}{a})
\\= sin^{-1}(\frac{t}{4})+C\\ =\sin^{-1}(\frac{x+3}{4})+C

Question:13 Integrate the functions \frac{1}{\sqrt { ( x-1)( x-2 )}}

Answer:

(x-1)(x-2) can be also written as
= x^2-3x+2
= (x-\frac{3}{2})^2-(\frac{1}{2})^2

therefore

\int \frac{1}{\sqrt{(x-1)(x-2)}}dx= \int \frac{1}{\sqrt{(x-\frac{3}{2})^2-(\frac{1}{2})^2}}dx
let suppose
x-3/2 = t \Rightarrow dx =dt
Now,

\Rightarrow \int \frac{1}{\sqrt{(x-\frac{3}{2})^2-(\frac{1}{2})^2}}dx = \int \frac{1}{\sqrt{t^2-(\frac{1}{2})^2}}dt .............by using formula \int \frac{1}{\sqrt{x^2-a^2}}=\log\left | x+\sqrt{x^2+a^2} \right |
\\= \log \left | t+\sqrt{t^2-(1/2)^2} \right |+C\\ = \log \left | (x-\frac{3}{2})+\sqrt{x^2-3x+2} \right |+C

Question:14 Integrate the functions \frac{1}{\sqrt { 8 + 3 x - x ^ 2 }}

Answer:

We can write denominator as
\\=8-(x^2-3x+\frac{9}{4}-\frac{9}{4})\\ =\frac{41}{4}-(x-\frac{3}{2})^2

therefore
\Rightarrow \int \frac{1}{\sqrt{8+3x-x^2}}dx= \int \frac{1}{\sqrt{\frac{41}{4}-(x-\frac{3}{2})^2}}
let x-3/2 = t \Rightarrow dx =dt

\therefore
\\=\int \frac{1}{\sqrt{(\frac{\sqrt{41}}{2})^2-t^2}}dt\\ =\sin^{-1}(\frac{t}{\frac{\sqrt{41}}{2}})+C\\ =\sin^{-1}(\frac{2x-3}{\sqrt{41}})+C

Question:15 Integrate the functions \frac{1}{\sqrt {(x-a)( x-b )}}

Answer:

(x-a)(x-b) can be written as x^2-(a+b)x+ab
\\x^2-(a+b)x+ab+\frac{(a+b)^2}{4}-\frac{(a+b)^2}{4}\\ (x-\frac{(a+b)}{2}^2)^2-\frac{(a-b)^2}{4}

\Rightarrow \int\frac{1}{\sqrt{(x-a)(x-b)}}dx=\int \frac{1}{\sqrt{(x-\frac{(a+b)}{2}^2)^2-\frac{(a-b)^2}{4}}}dx
let
x-\frac{(a+b)}{2}=t \Rightarrow dx =dt
So,
\\=\int \frac{1}{\sqrt{t^2-(\frac{a-b}{2})^2}}dt\\ =\log \left | t+\sqrt{t^2-(\frac{a-b}{2})^2} \right |+C\\ =\log \left | x-(\frac{a+b}{2})+\sqrt{(x-a)(x-b)} \right |+C

Question:16 Integrate the functions \frac{4x+1 }{\sqrt {2x ^ 2 + x -3 }}

Answer:

let
\\4x+1 = A\frac{d}{dx}(2x^2+x-3)+B\\ 4x+1=A(4x+1)+B\\ 4x+1=4Ax+A+B

By equating the coefficient of x and constant term on each side, we get
A = 1 and B=0

Let (2x^2+x-3) = t\Rightarrow (4x+1)dx =dt

\therefore \int \frac{4x+1}{\sqrt{2x^2+x-3}}dx= \int\frac{1}{\sqrt{t}}dt
\\= 2\sqrt{t}+C\\ =2\sqrt{2x^2+x-3}+C

Question:17 Integrate the functions \frac{x+ 2 }{\sqrt { x ^2 -1 }}

Answer:

let x+2 =A\frac{d}{dx}(x^2-1)+B=A(2x)+B
By comparing the coefficients and constant term on both sides, we get;

A=1/2 and B=2
then x+2 = \frac{1}{2}(2x)+2

\int \frac{x+2}{\sqrt{x^2-1}}dx =\int\frac{1/2(2x)+2}{x^2-1}dx
\\=\frac{1}{2}\int\frac{(2x)}{\sqrt{x^2-1}}dx+\int \frac{2}{\sqrt{x^2-1}}dx\\ =\frac{1}{2}[2\sqrt{x^2-1}]+2\log\left | x+\sqrt{x^2-1} \right |+C\\ =\sqrt{x^2-1}+2\log\left | x+\sqrt{x^2-1} \right |+C

Question:18 Integrate the functions \frac{5x -2 }{1+ 2x +3x^2 }

Answer:

let
\\5x+2 = A\frac{d}{dx}(1+2x+3x^2)+B\\ 5x+2= A(2+6x)+B = 2A+B+6Ax
By comparing the coefficients and constants we get the value of A and B

A = 5/6 and B = -11/3

NOW,
I = \frac{5}{6}\int \frac{6x+2}{3x^2+2x+1}dx-\frac{11}{3}\int \frac{dx}{3x^2+2x+1}
I = I_{1}-\frac{11}{3}I_{2} ...........................(i)

put 3x^2+2x+1 =t \Rightarrow (6x+2)dx =dt
Thus
I_{1}=\frac{5}{6}\int \frac{dt}{t} =\frac{5}{6}\log t =\frac{5}{6}\log (3x^2+2x+1)+c1
I_{2}= \int \frac{dx}{3x^2+2x+1} = \frac{1}{3}\int\frac{dx}{(x+1/3)^2+(\sqrt{2}/3)^2}
\\=\frac{1}{\sqrt{2}}\tan^{-1}(\frac{3x+1}{\sqrt{2}})+c2

\therefore I = I_1+I_2
I = \frac{5}{6}\log(3x^2+2x+1)-\frac{11}{3}\frac{1}{\sqrt{2}}\tan^{-1}(\frac{3x+1}{\sqrt2})+C

Question:19 Integrate the functions \frac{6x + 7 }{\sqrt {( x-5 )( x-4)}}

Answer:

let
6x+7 = A\frac{d}{dx}(x^2-9x+20)+B =A(2x-9)+B
By comparing the coefficients and constants on both sides, we get
A =3 and B =34

I =\int \frac{6x+7}{\sqrt{x^2-9x+20}}dx = \int \frac{3(2x+9)}{\sqrt{x^2-9x+20}}dx+34\int\frac{dx}{\sqrt{x^2-9x+20}} I = I_1+I_2 ....................................(i)

Considering I_1

I_1 =\int \frac{2x-9}{\sqrt{x^2-9x+20}}dx let x^2-9x+20 = t \Rightarrow (2x-9)dx =dt

I_1=\int \frac{dt}{\sqrt{t}} = 2\sqrt{t}=2\sqrt{x^2-9x+20}

Now consider I_2

I_2=\int \frac{dx}{\sqrt{x^2-9x+20}}
here the denominator can be also written as
Dr = (x-\frac{9}{2})^2-(\frac{1}{2})^2

\therefore I_2 = \int \frac{dx}{\sqrt{(x-\frac{9}{2})^2-(\frac{1}{2})^2}}
\\= \log\left | (x-\frac{9}{2})^2+\sqrt{x^2-9x+20} \right |

Now put the values of I_1 and I_2 in eq (i)

\\I = 3I_1+34I_2\\ I=6\sqrt{x^2-9x+20}+34\log\left | (x-\frac{9}{2})+\sqrt{x^2-9x+20} \right |+C

Question:20 Integrate the functions \frac{x +2 }{\sqrt { 4x - x ^ 2 }}

Answer:

let
x+2 = A\frac{d}{dx}(4x-x^2)+B = A(4-2x)+B
By equating the coefficients and constant term on both sides we get

A = -1/2 and B = 4

(x+2) = -1/2(4-2x)+4

\\\therefore \int \frac{x+2}{\sqrt{4x-x^2}}dx = -\frac{1}{2}\int \frac{4-2x}{\sqrt{4x-x^2}}+4\int \frac{dx}{\sqrt{4x-x^2}}\\ \ I =\frac{-1}{2}I_1+4I_2 ....................(i)

Considering I_1
\int \frac{4-2x}{\sqrt{4x-x^2}}dx
let 4x-x^2 =t \Rightarrow (4-2x)dx =dt
I_1=\int \frac{dt}{\sqrt{t}} = 2\sqrt{t}=2\sqrt{4x-x^2}
now, I_2

I_2 =\int \frac{dx}{\sqrt{4x-x^2}} = \int \frac{dx}{\sqrt{2^2-(x-2)^2}}
=\sin^{-1}(\frac{x-2}{2})

put the value of I_1 and I_2

I =-\sqrt{4x-x^2}+4\sin^{-1}(\frac{x-2}{2})+C

Question:21 Integrate the functions \frac{x +2 }{\sqrt { x^ 2 + 2x +3 }}

Answer:

\frac{x +2 }{\sqrt { x^ 2 + 2x +3 }}
\int \frac{x+2}{\sqrt{x^2+2x+3}}dx = \frac{1}{2}\int \frac{2(x+2)}{\sqrt{x^2+2x+3}}dx
\\= \frac{1}{2}\int \frac{2x+2}{\sqrt{x^2+2x+3}}dx+\frac{1}{2}\int \frac{2}{\sqrt{x^2+2x+3}}dx\\ =\frac{1}{2}\int \frac{2x+2}{\sqrt{x^2+2x+3}}dx+\int \frac{1}{\sqrt{x^2+2x+3}}dx\\ I=\frac{1}{2}I_1+I_2 ...........(i)

take I_1

\int \frac{2x+2}{\sqrt{x^2+2x+3}}dx
let x^2+2x+3 = t \Rightarrow (2x+2)dx =dt

I_1=\int \frac{dt}{\sqrt{t}}=2\sqrt{t}=2\sqrt{x^2+2x+3}
considering I_2

= \int \frac{dx}{\sqrt{x^2+2x+3}}= \int \frac{dx}{\sqrt{(x+1)^2+(\sqrt{2})^2}}
= \log \left | (x+1)+\sqrt{x^2+2x+3} \right |
putting the values in equation (i)

I=\sqrt{x^2+2x+3} +\log \left | (x+1)+\sqrt{x^2+2x+3} \right |+C

Question:22 Integrate the functions \frac{x + 3 }{x ^ 2 - 2x - 5 }

Answer:

Let (x+3) =A\frac{d}{dx}(x^2-2x+5)+B= A(2x-2)+B

By comparing the coefficients and constant term, we get;

A = 1/2 and B =4

\\\int \frac{x+3}{x^2-2x+5}dx = \frac{1}{2}\int \frac{2x-2}{x^2-2x+5}dx +4\int \frac{1}{x^2-2x+5}dx\\ I=I_1+I_2 ..............(i)

\\\Rightarrow I_1\\ =\int \frac{2x-2}{x^2-2x-5}dx
put x^2-2x-5 =t \Rightarrow (2x-2)dx =dt

=\int \frac{dt}{t} = \log t = \log (x^2-2x-5)

\\\Rightarrow I_2\\ = \int \frac{1}{x^2-2x-5}dx\\ =\int \frac{1}{(x-1)^2+(\sqrt{6})^2}dx\\ =\frac{1}{2\sqrt{6}}\log(\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}})

I=I_1+I_2

=\frac{1}{2}\log\left | x^2-2x-5 \right |+\frac{2}{\sqrt{6}}\log(\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}})+C

Question:23 Integrate the functions \frac{5x + 3 }{\sqrt { x^2 + 4x +10 }}

Answer:

let
5x+3 = A\frac{d}{dx}(x^2+4x+10)+B = A(2x+4)+B
On comparing, we get

A =5/2 and B = -7

\int \frac{5x+3}{\sqrt{x^2+4x+10}}dx = \frac{5}{2}\int \frac{2x+4}{\sqrt{x^2+4x+10}}dx-7\int \frac{dx}{\sqrt{x^2+4x+10}}dx I = 5/2I_1-7I_2 ...........................................(i)

\\\Rightarrow I_1\\ \int \frac{2x+4}{\sqrt{x^2+4x+10}}dx
put
x^2+4x+10= t \Rightarrow (2x+4)dx = dt

=\int \frac{dt}{\sqrt{t}}=2\sqrt{t}=2\sqrt{x^2+4x+10}

\\\Rightarrow I_2\\ =\int \frac{1}{\sqrt{x^2+4x+10}}dx \\ =\int \frac{1}{\sqrt{(x+2)^2+(\sqrt{6})^2}}dx\\ =\log \left | (x+2)+\sqrt{x^2+4x+10} \right |

I = 5\sqrt{x^2+4x+10}-7\log\left | (x+2)+\sqrt{x^2+4x+10} \right |+C

Question:24 Choose the correct answer

\int \frac{dx }{x^2 + 2x +2 }\: \: equals

(A) x \tan^{-1} (x + 1) + C\\\\ (B) \tan^{-1} (x + 1) + C\\\\ (C) (x + 1) \tan^{-1}x + C \\\\ (D) \tan^{-1}x + C

Answer:

The correct option is (B)

\int \frac{dx }{x^2 + 2x +2 }\: \: equals
the denominator can be written as (x+1)^2+1
now, \int \frac{dx}{(x+1)^2+1} = tan^{-1}(x+1)+C

Question:25 Choose the correct answer \int \frac{dx }{\sqrt { 9x - 4x ^2 }} \: \: equals

A) \frac{1}{9} \sin ^{-1}\left ( \frac{9x-8}{8} \right )+ C \\\\B ) \frac{1}{2} \sin ^{-1}\left ( \frac{8x-9}{9} \right )+ C \\\\ C) \frac{1}{3} \sin ^{-1}\left ( \frac{9x-8}{8} \right )+ C \\\\ D ) \frac{1}{2} \sin ^{-1}\left ( \frac{9x-8}{8} \right )+ C

Answer:

The following integration can be done as

\int \frac{dx }{\sqrt { 9x - 4x ^2 }} \: \: equals
\int \frac{1}{\sqrt{-4(x^2-\frac{9}{4}x)}}= \int \frac{1}{\sqrt{-4(x^2-\frac{9}{4}x+81/64-81/64)}}dx
\\= \int \frac{1}{\sqrt{-4[(x-9/8)^2-(9/8)^2]}}dx\\ =\frac{1}{2}\int \frac{1}{\sqrt{-(x-9/8)^2+(9/8)^2}}dx\\ =\frac{1}{2}[\sin^{-1}(\frac{x-9/8}{9/8})]+C\\ =\frac{1}{2}\sin^{-1}(\frac{8x-9}{9})+C

The correct option is (B)



More About NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.4

The NCERT Class 12 Maths chapter Integrals is a good source to cover integrals from basics to advanced level. Exercise 7.4 Class 12 Maths can help students to get an idea of types of questions asked in the examination. Overall NCERT solutions for Class 12 Maths chapter 7 exercise 7.4 is a good source to practice before the exam.

Also Read| Integrals Class 12 Notes

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