∫ x^3 dx / 1 + x^8 =

Question

\(\int \frac{x^{3} \mathrm{~d} x}{1+x^{8}}=\)

(A) \(\tan ^{-1} x^{4}+c\)

(B) \(4 \tan ^{-1} x^{4}+c\)

(C) \(\frac{1}{4} \tan ^{-1} x^{4}+c\)

(D) \(2 \tan ^{-1} x^{4}+c\)

Answer 1

Correct option is (C) \(\frac{1}{4} \tan ^{-1} x^{4}+c\)

\(I = \int \frac{x^{3} \mathrm{~d} x}{1+x^{8}}\)

\(I = \int \frac{x^{3}\mathrm{~d} x}{1+(x^{4})^2} \)

Put \(x^4 = t\) and \(4x^3 dx = dt\)

\(I = \frac 14 \int \frac{dt}{1 + t^2}\)

\(= \frac 14 \tan^{-1}t + c\)

\(= \frac 14 \tan^{-1}x^4+ c\)