First slide

Evaluation of definite integrals

Question

$\int_{- 1}^{1} \frac{x^{3} + | x | + 1}{x^{2} + 2 | x | + 1} dx$ is equal to

Moderate

A

log 2
B

2 log 2
C

$\frac{1}{2}$ log 2
D

4 log 2

Solution

Since,
$\begin{array}{l} I = \int_{- 1}^{1} \frac{x^{3} + | x | + 1}{x^{2} + 2 | x | + 1} dx \\ I = \int_{- 1}^{1} \frac{x^{3}}{x^{2} + 2 | x | + 1} + \int_{- 1}^{1} \frac{| x | + 1}{x^{2} + 2 | x | + 1} dx \\ = 0 + 2 \int_{0}^{1} \frac{| x | + 1}{(| x | + 1)^{2}} dx [odd function + even function] \\ = 2 \int_{0}^{1} \frac{x + 1}{(x + 1)^{2}} dx = 2 \int_{0}^{1} \frac{1}{x + 1} dx \\ = 2 [\log ∣ x + 1 ∥]_{0}^{1} = 2 \log 2 \end{array}$

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