Practicing Success
Practicing Success
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$\int \frac{\left(\tan ^{-1} x\right)^3}{1+x^2} d x$ is equal to |
$3\left(\tan ^{-1} x\right)^2+C$ $\frac{\left(\tan ^{-1} x\right)^4}{4}+C$ $\left(\tan ^{-1} x\right)^4+C$ none of these |
$\frac{\left(\tan ^{-1} x\right)^4}{4}+C$ |
We have, $\int \frac{\left(\tan ^{-1} x\right)^3}{1+x^2} d x=\int\left(\tan ^{-1} x\right)^3 d\left(\tan ^{-1} x\right)=\frac{\left(\tan ^{-1} x\right)^4}{4}+C$ |
