Let f(x)=\;{x² dx}{(1+x²)(1+{1+x²)} and f(0)=0, then the value of f(1) be

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

Test Index

BITSAT 2022 Paper

Section: Mathematics

Question : 119 of 130

Marks: +1, -0

Let

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

f(0)=0

f(1)

Solution: 👈: Video Solution

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

Let

x = \tan\theta

\Rightarrow dx = \sec^2\theta d\theta = (1+x^2)d\theta

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

= \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(1+\cos\theta)}

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

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

= \log(x+\sqrt{1+x^2}) - \tan^{-1}x + C

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

0 = \log 1 - 0 + C

\Rightarrow C = 0

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

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

Go to Question:

Prev Question

Next Question