AP EAMCET · Maths · Indefinite Integration
\(\int \frac{x^5}{x^2+1} d x=\)
- A \(\frac{x^4}{4}+\frac{x^3}{3}-\tan ^{-1} x+c\)
- B \(\frac{x^4}{4}-\frac{x^2}{2}+\frac{1}{2} \log \left(x^2+1\right)+c\)
- C \(\frac{x^4}{4}+\frac{x^3}{3}+\tan ^{-1} x+c\)
- D \(\frac{x^4}{4}+\frac{x^2}{2}-\frac{1}{2} \log \left(x^2+1\right)+c\)
Answer & Solution
Correct Answer
(B) \(\frac{x^4}{4}-\frac{x^2}{2}+\frac{1}{2} \log \left(x^2+1\right)+c\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { } I=\int \frac{x^5}{x^2+1} d x=\int \frac{\left(x^4-1+1\right) x d x}{\left(x^2+1\right)} \\ & =\int\left(x^2-1+\frac{1}{x^2+1}\right) d x=\frac{x^4}{4}-\frac{x^2}{2}+\frac{1}{2} \log \left(x^2+1\right)+c\end{aligned}\)
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