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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Integral Calculus

Integrate : $\int \large\frac{7}{9} \times \frac{e^{\Large \tan ^{-1}(x^4)}}{(1+x^8)} $$\times x^3 dx$

$(a)\;\tan ^{-1}(x^4)+c \qquad(b)\;\cot^{-1}(x^4)+c \qquad(c)\;\frac{7}{36} e^{\cos^{-1}(x^4)}+c \qquad (d)\;\frac{7}{36}e^{\tan ^{-1}(x^4)}+c$

1 Answer

$\tan ^{-1} (x^4) =t$
differentiate with respect to x
$\large\frac{1}{(1+x^8)} $$ \times 4x^3 dx=dt$
=>$\large\frac{7}{9} \int \frac{1}{4} $$e^t dt$
=> $\large\frac{7}{36} $$e^t +c$
=> $\large\frac{7}{36} e^{\tan ^{-1} (x^4)} +c$
Hence d is the correct answer.
answered Dec 24, 2013 by meena.p
 
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