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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Integral Calculus
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Integrate: $ \int \large\frac{x^3 \tan ^{-1} x^4 }{1+x^8}$$dx$

$(a)\;\frac{[\tan ^{-1} (x^4)]^2}{8}+c \qquad(b)\; \frac{[\cot ^{-1} (x^4)]^2}{8}+c\qquad(c)\;\cot ^{-1}x^4+c \qquad (d)\;None$

Can you answer this question?
 
 

1 Answer

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Suppose : $\tan ^{-1} (x^4)=t$
Differentiate with (x)
=> $\large\frac{1}{1+ (x^4)^2} \times $$4x^3 dx=dt$
=> $\large\frac{x^3}{1+x^8} $$dx$
=>$\large\frac{dt}{4}$-----(1)
$\qquad= \int \large\frac{t}{4} $$dt$
$\qquad =\large\frac{t^2}{2 \times 4}$$+c$
$\qquad =\large\frac{t^2}{8}$$+c$
=>$\large\frac{[\tan ^{-1} (x^4)]^2}{8}$$+c$
Hence a is the correct answer.

 

answered Dec 23, 2013 by meena.p
edited Mar 20, 2014 by balaji.thirumalai
 
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