Definite Integration Question 70
Question 70
- Prove that
Hence or otherwise, evaluate the integral
Show Answer
Solution:
Now,
where, $I_{1}=\int_{0}^{1} \tan ^{-1} x d x=\left[x \tan ^{-1} x\right]{0}^{1}-\int{0}^{1} \frac{x d x}{1+x^{2}}$
$$ \begin{gathered} =\frac{\pi}{4}-\frac{1}{2}\left[\log \left(1+x^{2}\right)\right]{0}^{1}=\frac{\pi}{4}-\frac{1}{2} \log 2 \ \therefore \int{0}^{1} \tan ^{-1}\left(1-x+x^{2}\right) d x=\frac{\pi}{2}-2 \frac{\pi}{4}-\frac{1}{2} \log 2=\log 2 \end{gathered} $$
