Question

If $f\left(x\right)=\int \left(\frac{x^2+{\sin }^{2}x}{1+x^2}\right){\sec }^{2}xdx$ and f(0)=0,then f (1) equals:

• A. $1-\frac{\pi }{4}$
• B. $-\frac{\pi }{4}$
• C. $\tan 1-\frac{\pi }{4}$
• D. $\tan 1+1$

Answer 1

The correct option is C $\tan 1-\frac{\pi }{4}$

$f\left(x\right)=\int \left(\frac{x^2+si{n}^{2}x}{1+x^2}\right)se{c}^{2}xdx$

$=\int \left(\frac{x^2+1-(1-si{n}^{2}x)}{1+x^2}se{c}^{2}x\right)dx$

$=\int (se{c}^{2}x-\frac{co{s}^{2}xse{c}^{2}x}{1+x^2})dx$
$=\int se{c}^{2}x-\int \frac{1}{1+x^2}dx$

$f\left(x\right)=tanx-ta{n}^{-1}x+\alpha$

$f\left(0\right)=0\Rightarrow \alpha =0$

$f\left(1\right)=tan1-\frac{\pi }{4}$