$\int_{0}^{5} \frac{\mathrm{d} x}{x^{2}+2 x+10}$ =

$\int_{0}^{5} \frac{\mathrm{d} x}{x^{2}+2 x+10}$ =
  1. $\frac{\pi}{6}$
  2. $\frac{\pi}{12}$
  3. $\frac{\pi}{3}$
  4. $\frac{\pi}{4}$

Solution

$\begin{aligned} I &=\int_{0}^{5} \frac{d x}{x^{2}+2 x+10}=\int_{0}^{5} \frac{d x}{(x+1)^{2}+(3)^{2}} \\ &=\frac{1}{3}\left[\tan ^{-1} \frac{x+1}{3}\right]_{0}^{5}=\frac{1}{3}\left[\tan ^{-1} 2-\tan ^{-1} \frac{1}{3}\right] \\ &=\frac{1}{3}\left[\tan ^{-1}\left[\frac{2-\frac{1}{3}}{1+\frac{2}{3}}\right]\right]=\frac{1}{3} \tan ^{-1}(1) \\ &=\frac{1}{3} \times \frac{\pi}{4}=\frac{\pi}{12} \end{aligned}$

Asked in: MHT CET 2020 (15 Oct Shift 2)

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