If f(x)= {x² \, dx}{(1+x²)(1+{1+x²)} and f(0)=0, then f(1) is

Correct Answer is : log⁡(1+2)−π4 (1+{2}) - π/4log(1+2)−4π

Correct Answer is : log⁡(1+2)−π4 (1+{2}) - π/4log(1+2)−4π

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MHTCET 13 May 2023 Shift 1 Solved Paper

Section: Mathematics

Question : 76 of 150

Marks: +1, -0

f(x)=\int \frac{x^2 \, dx}{(1+x^2)(1+\sqrt{1+x^2})}

f(0)=0

f(1)

Solution:

\text{Put } x=\tan\theta \Rightarrow dx=\sec^2\theta \, d\theta

\therefore\;\; f(x)=\int \frac{\tan^2\theta \sec^2\theta \, d\theta}{\sec^2\theta(1+\sec\theta)}

=\int \frac{\tan^2\theta \, d\theta}{1+\sec\theta}

=\int \frac{\sin^2\theta \, d\theta}{\cos\theta(1+\cos\theta)}

=\int \frac{1-\cos^2\theta \, d\theta}{\cos\theta(1+\cos\theta)}

=\int \frac{(1-\cos\theta) \, d\theta}{\cos\theta}

=\int \sec\theta \, d\theta - \int d\theta

=\log |\sec\theta+\tan\theta| - \theta + c

\therefore\;\; f(x)=\log |x+\sqrt{1+x^2}| - \tan^{-1} x + c

\therefore f(0)=\log |0+\sqrt{1+0}| - \tan^{-1}(0) + c

\Rightarrow\;\; 0=\log 1 - 0 + c \Rightarrow c=0

\therefore f(x)=\log |x+\sqrt{1+x^2}| - \tan^{-1} x

\therefore f(1)=\log |1+\sqrt{1+1^2}| - \tan^{-1}(1)

=\log (1+\sqrt{2}) - \frac{\pi}{4}

\;

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