The integral _0^1/2 (1+2x)1+4x^2,dx, equals:

Question

The integral $\displaystyle\int_0^{1/2} \dfrac{\ln(1+2x)}{1+4x^2}\,dx$, equals:

Accepted Answer

Put $2x=	an	heta \Rightarrow dx = \dfrac{\sec^2	heta}{2}\,d	heta$; limits $x=0\Rightarrow	heta=0$, $x=\dfrac12\Rightarrow	heta=\dfrac{\pi}{4}$

$I = \dfrac12\int_0^{\pi/4} \ln(1+	an	heta)\,d	heta$ (i)

Using $\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$:

$I = \dfrac12\int_0^{\pi/4}\ln\left(1+\dfrac{1-	an	heta}{1+	an	heta}ight)d	heta = \dfrac12\int_0^{\pi/4}\ln\left(\dfrac{2}{1+	an	heta}ight)d	heta$

$= \dfrac12\int_0^{\pi/4}[\ln 2 - \ln(1+	an	heta)]\,d	heta = \dfrac12\ln2\cdot\dfrac{\pi}{4} - I$

$2I = \dfrac{\pi}{8}\ln2 \Rightarrow I = \dfrac{\pi}{16}\ln2$

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