Question

Integrate Log(1+x)/ 1+x2 dx

Upper Limit 1

Lower limit 0

Answer 1

your question is -> $\int\limits^1_0{\frac{log(1+x)}{(1+x^2)}}\,dx$

you can easily solve this question with help of substitution method

put x = tanθ

differentiating both sides,

dx = sec²θ dθ

now limits : upper limits = 1 = tanθ ⇒θ = π/4 and lower limits = 0 = tanθ ⇒θ = 0°

now, $\int\limits^1_0{\frac{log(1+x)}{(1+x^2)}}\,dx=\int\limits^{\pi/4}_0{\frac{log(1+tan\theta)}{(1+tan^2\theta)}}sec^2\theta d\theta$

I = $\int\limits^{\pi/4}_0{\frac{log(1+tan\theta)}{(sec^2\theta)}}sec^2\theta d\theta$

I = $\int\limits^{\pi/4}_0{log(1+tan\theta)}\,d\theta$.....(1)

we know, $\int\limits^a_0f(x)\,dx=\int\limits^a_0f(a-x)\,dx$

so, $I=\int\limits^{\pi/4}_0{log(1+tan\theta)}\,d\theta=\int\limits^{\pi/4}_0{log(1+tan(\pi/4-\theta))}\,d\theta$

we know, tan(π/4- θ) = (1 - tanθ)/(1 + tanθ)

so, I = $\int\limits^{\pi/4}_0{log\left(1+\frac{1-tan\theta}{1+tan\theta}\right)}\,d\theta$

= $\int\limits^{\pi/4}_0{log\frac{2}{1+tan\theta}}\,d\theta$

= $\int\limits^{\pi/4}_0{\left[log2-log(1+tan\theta)\right]}\,d\theta$......(2)

adding equations (1) and (2),

we get, 2I = $\int\limits^{\pi/4}_0log2$

so, I = $\frac{\pi}{8}log2$