Question

$\int \frac{(\cos x+x\sin x)}{(x^2+x\cos x)}dx=$

• A. $\log \left|\frac{(\sin x)}{(1+\cos x)}\right|+c$
• B. $\log \left|\frac{(\sin x)}{(x+\cos x)}\right|+c$
• C. $\log \left|\frac{(2\sin x)}{(x+\cos x)}\right|+c$
• D. $\log \left|\frac{(x\sin x)}{(x+\cos x)}\right|+c$
• E. $\log \left|\frac{\left(x\right)}{(x+\cos x)}\right|+c$

Answer 1

The correct option is E

$\log \left|\frac{\left(x\right)}{(x+\cos x)}\right|+c$

Explanation for the correct option:

Evaluate the given integral:

$\int \frac{\cos x+x\sin x}{x^2+x\cos x}dx$

Adding and subtracting $x$ in the numerator

$$\begin{gathered} \int \frac{\cos x+x\sin x+x-x}{x^2+x\cos x}dx \\ =\int \frac{x+\cos x}{x^2+x\cos x}dx-\int \frac{x-x\sin x}{x^2+x\cos x} \\ =\int \frac{x+\cos x}{x\left(x+\cos x\right)}dx-\int \frac{x-x\sin x}{x\left(x+\cos x\right)}dx \\ =\int \frac{dx}{x}-\int \frac{x-x\sin x}{x\left(x+\cos x\right)}dx \\ =\log \left|x\right|-\int \frac{x\left(1-\sin x\right)}{x\left(x+\cos x\right)}dx \\ =\log \left|x\right|-\int \frac{\left(1-\sin x\right)}{\left(x+\cos x\right)}dx \end{gathered}$$

Let $$\begin{gathered} z=x+\cos x \\ \Rightarrow dz=\left(1-\sin x\right)dx \end{gathered}$$

$$\begin{gathered} =\log \left|x\right|-\int \frac{dz}{z} \\ =\log \left|x\right|-\log \left|z\right|+c \\ =\log \left|x\right|-\log \left|x+\cos x\right|+c\left[\text{substituting the value of}z\right] \\ =\log \left|\frac{x}{x+\cos x}\right|+c\left[\because \log a-\log b=\log \left|\frac{a}{b}\right|\right] \end{gathered}$$

Therefore, $\int \frac{(\cos x+x\sin x)}{(x^2+x\cos x)}dx=$$\log \left|\frac{\left(x\right)}{(x+\cos x)}\right|+c$

Hence, the correct answer is option (E).