\int \frac{x^2 \tan ^{-1} x}{ (1+x^2) ^2} \dx=

Correct Answer is : (tan−1x)24−4xtan−1x+1−x28(1+x2)+C

\frac{ (\tan ^{-1} x) ^2}{4}-\frac{x \tan ^{-1} x}{2 (1+x^2) }+\frac{1-x^2}{4 (1+x^2) }+C

\frac{ (\tan ^{-1} x) ^2}{4}-\frac{4 x \tan ^{-1} x+1-x^2}{8 (1+x^2) }+C

\frac{ (\tan ^{-1} x) ^2}{4}-\frac{x \tan ^{-1} x}{ (1+x^2) }-\frac{1-x^2}{4 (1+x^2) }+C

\frac{(\tan x)^2}{4}+\frac{4 x \tan ^{-1} x-1+x^2}{4 (1+x^2) }+C

TG EAMCET 3 May 2025 Shift 1 Paper

Question : 74 of 160

Marks: +1, -0

∫‌

x2tan−1x

(1+x2)2

‌dx=

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