# Evaluate : $\int \large\frac{3x+5}{\sqrt{x^2-8x+7}}$$dx ## 1 Answer Toolbox: • \int\large\frac{(px+q)}{\sqrt{ax^2+bx+c}}$$dx.$,where p,q,a,b,c are constants,we are to find real numbers such that $px+q=A\frac{d}{dx}(ax^2+bx+c)+B.$
• Therefore $px+q=A(2ax+b)+B.$
• To determine A and B we equate the coefficients from both sided,the coefficients of x and the constant terms.A and B are thus obtained and hence the integral is reduced to one of the known forms:
• $\int\large\frac{dx}{\sqrt{x^2+a^2}}$$=\log\mid x+\sqrt{x^2+a^2}\mid+c. Step 1: Let I=\int\large\frac{3x+5}{\sqrt{x^2-8x+7}}$$dx$
$3x+5=A\large\frac{d}{dx}$$(x^2-8x+7)+B 3x+5=A(2x-8)+B Equating the coefficients of x we get, 3=2A\Rightarrow A=\large\frac{3}{2} Equating the constant terms, 5=-8\big(\large\frac{3}{2}\big)$$+B$
$\Rightarrow B=17$
$\therefore A=\large\frac{3}{2}$ and $B=17$
Step 2:
Now substituting the values of $A$ and $B$ we get,
$I=\int\large\frac{3/2(2x-8)}{\sqrt{x^2-8x+7}}$$dx+17\int\large\frac{dx}{\sqrt{x^2-8x+7}} I=\large\frac{3}{2}$$I_1+17I_2$
Consider $I_1$
Put $x^2-8x+7=t$
$\Rightarrow (2x-8)dx=dt$
$\therefore I_1=\int\large\frac{dt}{\sqrt t}$
$\Rightarrow \sqrt t+c_1$
$\Rightarrow \sqrt{x^2-8x+7}+c$
Step 3:
Consider $I_2=\int\large\frac{dx}{\sqrt{x^2-8x+7}}$
$\Rightarrow \int\large\frac{dx}{\sqrt{x-4)^2-16+7}}$
$I_2=\int\large\frac{dx}{(x-4)^2-9}$
$\;\;=\int\large\frac{dx}{\sqrt{(x-4)^2-(3)^2}}$
$\;\;=\log\mid(x-4)+\sqrt{x^2-8x+7}\mid+c_2$
Step 4:
Combining the terms we get,