If f(x) = x^2+^2x1+x^2^2x,dx and f(0)=0, then f(1) equals:

Question

If $f(x) = \displaystyle\int \dfrac{x^2+\sin^2x}{1+x^2}\sec^2x\,dx$ and $f(0)=0$, then $f(1)$ equals:

Accepted Answer

$f(x) = \int \left(x^2\sec^2x+\dfrac{\sin^2x}{\cos^2x(1+x^2)}ight)dx$ --- reorganizing:

$= \int \dfrac{x^2\sec^2x(1+x^2)+	an^2x}{1+x^2}\,dx$, but simplifying as in source:

$= \int \dfrac{x^2(1+	an^2x)+	an^2x}{1+x^2}\,dx = \int \dfrac{x^2+	an^2x(1+x^2)}{1+x^2}\,dx$

$= \int \dfrac{x^2}{1+x^2}\,dx + \int 	an^2x\,dx = \int\left(1-\dfrac{1}{1+x^2}ight)dx + \int(\sec^2x-1)\,dx$

$= x - 	an^{-1}x + 	an x - x + c = 	an x - 	an^{-1}x + c$

$f(0)=0 \Rightarrow c=0$

$f(1) = 	an1-	an^{-1}1 = 	an1-\dfrac{\pi}{4}$

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