Evaluate the integral _ -1 ^ 1 d x x^ 2 +2 x+5 using substitution…

Question

Evaluate the integral $\int_{-1}^{1} \frac{d x}{x^{2}+2 x+5}$ using substitution.

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Answer

Given: $\int_{-1}^{1} \frac{d x}{x^{2}+2 x+5}$
= $\int_{-1}^{1} \frac{d x}{\left(x^{2}+2 x+1\right)+4}$
= $\int_{-1}^{1} \frac{d x}{(x+1)^{2}+(2)^{2}}$
Let x + 1 = t
⇒ dx = dt
When x = -1, t = 0 and when x = 1, t = 2
$\Rightarrow \int_{-1}^{1} \frac{d x}{(x+1)^{2}+(2)^{2}}=\int_{0}^{2} \frac{d t}{(t)^{2}+(2)^{2}}$
because, $\int \frac{d t}{x^{2}+a^{2}}=\frac{1}{a} \cdot \tan ^{-1} \frac{x}{a}+C$
$\Rightarrow \int_{0}^{2} \frac{d t}{(t)^{2}+(2)^{2}}=\left[\frac{1}{2} \tan ^{-1} \frac{t}{2}\right]_{0}^{2}$
= $\frac{1}{2} \tan ^{-1} 1-\frac{1}{2} \tan ^{-1} 0$
= $\frac{1}{2}\left(\frac{\pi}{4}\right)=\frac{\pi}{8}$

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