AP EAMCET · Maths · Indefinite Integration
\(\int \frac{x}{\left(x^2+2 x+2\right)^2} d x\) is equal to
- A \(\frac{x^2+2}{x^2+2 x+2}-\frac{1}{2} \tan ^{-1}(x+1)+C\)
- B \(\frac{x^2+2}{2\left(x^2+2 x+2\right)}-\frac{1}{2} \tan ^{-1}(x-1)+C\)
- C \(\frac{x^2-2}{4\left(x^2+2 x+2\right)}-\frac{1}{2} \tan ^{-1}(x+1)+C\)
- D \(\frac{2(x-1)}{\left(x^2+2 x+2\right)}+\frac{1}{2} \tan ^{-1}(x+1)+C\)
Answer & Solution
Correct Answer
(C) \(\frac{x^2-2}{4\left(x^2+2 x+2\right)}-\frac{1}{2} \tan ^{-1}(x+1)+C\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { Let, } l=\int \frac{x}{\left(x^2+2 x+2\right)^2} d x \\ & l=\int \frac{x}{\left[(x+1)^2+1\right]^2} d x \end{aligned}\) Put \(x+1=\tan \theta\)…
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