₀¹ _e (1+x)/1+x² dx =

Correct Answer is : π8log⁡eπ/8 _e8πloge 2

Correct Answer is : π8log⁡eπ/8 _e8πloge 2

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AP EAMCET Engineering 20 Apr 2019 Shift 1 Paper

Section: Mathematics

Question : 77 of 160

Marks: +1, -0

\int\limits_{0}^{1} \frac{\log_e (1+x)}{1+x^2}

dx =

Solution:

Consider the given integral.

I=\int\limits_{0}^{1} \; \frac{\log_e (1+x)}{1+x^{2}} \, dx

Take

x=\tan \theta,

then

dx=\sec^{2} \theta \, d\theta

Substitute in the above integral and change the limits accordingly.

I \;\; = \int\limits_{0}^{\frac{\pi}{4}} \frac{\log_e (1+\tan \theta)}{\sec^{2} \theta} \sec^{2} \theta \, d\theta

\;\; = \int\limits_{0}^{\frac{\pi}{4}} \log_e (1+\tan \theta) \, d\theta

Now, using the property

\int\limits_{0}^{a} f(x) \, dx = \int\limits_{0}^{a} f(a-x) \, dx

I \;\; = \int\limits_{0}^{\frac{\pi}{4}} \log_e \left(1+\tan \left(\frac{\pi}{4}-\theta\right)\right) \, d\theta

= \;\; \int\limits_{0}^{\frac{\pi}{4}} \log_e \left(1+\frac{1-\tan \theta}{1+\tan \theta}\right) \, d\theta

= \;\; \int\limits_{0}^{\frac{\pi}{4}} \log_e \left(\frac{2}{1+\tan \theta}\right) \, d\theta

= \;\; \int\limits_{0}^{\frac{\pi}{4}} \log_e (2) \, dx - \int\limits_{0}^{\frac{\pi}{4}} \log_e (1+\tan \theta) \, d\theta

Further simplifying, we get

I = \frac{\pi}{4} \log_e (2) - I

2I = \frac{\pi}{4} \log_e (2)

I = \frac{\pi}{8} \log_e (2)

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