Find the value of: $ \int _{0}^{ \infty} \frac{ \ln x}{x^2+2x+4}\,\text{d}x$
Here I factorised the denominator into complex factors, and performing partial fraction decomposition, I get the following integral I cannot solve: $\int \frac{\ln(x-a)}{x}\,\text{d}x$
where $a$ is a constant.
So how do I evaluate this, or is there any other way to approach the original definite integral?
Let $x=2y$, the the required integral is $$I=\int_{0}^{\infty} \frac{2\ln(2y) dy}{4(y^2+y+1)}=\int_{0}^{\infty} \frac{\ln 2 dy}{2(y^2+y+1)}+\int_{0}^{\infty} \frac{\log y dy}{2(y^2+y+1)}=K+J.~~~(1)$$ In the second integral call it $J$, let $y=1/t\Rightarrow dy=-dt/t^2$ then $$J=-\int_{0}^{\infty} \frac{\ln t dt}{2(t^2+t+1)} \Rightarrow J=-J \Rightarrow J=0.$$ The first integral in (1) call it $K$ is $$K=\int_{0}^{\infty} \frac{ \ln 2 dy}{2[(y+1/2)^2+3/4]}=\frac{\ln 2}{2} \int_{1/2}^{\infty} \frac{du}{u^2+(\sqrt{3}/2)^2}= \frac{\ln 2}{2} \frac{2}{\sqrt{3}}(\pi/2-\pi/6).$$ $$ \Rightarrow I =\frac{\pi \ln 2}{3 \sqrt{3}}.$$
$x+1=\sqrt3\tan t$
$$\sqrt3I=\int_{\pi/6}^{\pi/2}\ln(\sqrt3\tan t-1)dt$$
$$=\int_{\pi/6}^{\pi/2}(\ln2-\ln(\cos t)+\ln\sin(t-\pi/6))\ dt$$
Like Evaluate the integral $\int^{\frac{\pi}{2}}_0 \frac{\sin^3x}{\sin^3x+\cos^3x}\,\mathrm dx$.,
if $f(t)=\cos t, f(\pi/2+\pi/6-t)=?$
$$\implies\sqrt3I=\ln2\int_{\pi/6}^{\pi/2}dt$$
When you get limits as reciprocal of each other one of easier tool is take substitution let x=a/t. . In the given integral take x= 4/t You will get integration as ( dt / (t+1)^2 + 3 ) . Hence you will get the answer
