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If $\mathrm{f}(\mathrm{x})=\int \frac{\mathrm{x}^2+\sin ^2 \mathrm{x}}{1+\mathrm{x}^2}, \sec ^2 \mathrm{x} d \mathrm{x}$ and $\mathrm{f}(0)=0$, then $\mathrm{f}(1)=$
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Verified Answer
The correct answer is:
$\tan 1-\frac{\pi}{4}$
$\begin{aligned} & \mathrm{f}(\mathrm{x})=\int \frac{\mathrm{x}^2+\sin ^2 \mathrm{x}}{1+\mathrm{x}^2} \cdot \sec ^2 \mathrm{x} d \mathrm{x} \\ & =\int \frac{1+\mathrm{x}^2+\sin ^2 \mathrm{x}-1}{1+\mathrm{x}^2} \cdot \sec ^2 \mathrm{x} d \mathrm{x} \\ & =\int\left(1-\frac{\cos ^2 \mathrm{x}}{1+\mathrm{x}^2}\right) \sec ^2 \mathrm{x} d \mathrm{x} \\ & =\int \sec ^2 \mathrm{x} d \mathrm{x}-\int \frac{\mathrm{dx}}{1+\mathrm{x}^2} \\ & =\mathrm{f}(\mathrm{x})=\tan \mathrm{x}-\tan ^{-1} \mathrm{x}+\mathrm{C} \\ & \because \mathrm{f}(0)=0 \Rightarrow \mathrm{c}=0\end{aligned}$
So, $\mathrm{f}(\mathrm{x})=\tan \mathrm{x}-\tan ^{-1} \mathrm{X}$
$\Rightarrow \mathrm{f}(1)=\tan 1-\tan ^{-1}(1)=\tan (1)-\frac{\pi}{4}$
So, $\mathrm{f}(\mathrm{x})=\tan \mathrm{x}-\tan ^{-1} \mathrm{X}$
$\Rightarrow \mathrm{f}(1)=\tan 1-\tan ^{-1}(1)=\tan (1)-\frac{\pi}{4}$
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