Question

Evaluate in terms of natural logarithms
${\int }_4^7\frac{1}{(4x^2-8x-5)}dx$

Answer 1

The integral can be written as
${\int }_4^7\frac{1}{(4{(x-1)}^2-9)}dx$
You need to make use of the identity $\cos h^{2}A-1=\sin h^{2}A$ because of the appearance of the denominator
Substitute $4{(x-1)}^2=9\cos h^{2}u$ in order to accomplish this.
So, $2(x-1)=3\cos hu$
and $2\frac{dx}{du}=3\sin hx$
The denominator then becomes
$\{9\cos h^{2}u-9\}=\left(9\sin h^{2}u\right)=3\sin hu$
In order to deal with the limits, note that when
$x=4,\cos hu(sou=\ln [2+\sqrt{3}])$
$x=7,\cos hu(sou=\ln [4+\sqrt{15}])$
The integral then becomes
${\int }_{\cos h^{-1}2}^{\cos h^{-1}4}\frac{\sin hudu}{3\sin hu}={\int }_{\cos h^{-1}2}^{\cos h^{-1}4}\frac{1}{2}du=\frac{1}{2}[\cos h^{-1}4-\cos h^{-1}2]=\frac{1}{2}\{\ln (2+\sqrt{3})-\ln 4+\sqrt{15}\}$