# Integrate the functions$\frac{5x-2}{1+2x+3x^2}$

Toolbox:
• $(i)\int\frac{(px+q)}{\sqrt{ax^2+bx+c}}dx.$,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:
• (ii)$\int\frac{dx}{\sqrt{x^2+a^2}}=\frac{1}{a}\tan^{-1}\big(\frac{x}{a}\big)+c.$
Given:$I=\int \frac{5x-2}{1+2x+3x^2}dx.$

Let $5x-2=A\frac{d}{dx}(1+2x+3x^2)+B.$

5x-2=A(2+6x)+B.

Equating the coefficients we get,

5=6A $\Rightarrow A=\frac{5}{6}.$

-2=2A+B.

$-2=2\big(\frac{5}{6}\big)+B\Rightarrow B=\frac{-11}{3}.$

Hence $I=\int\frac{5/6(2+6x)-11/3}{1+2x+3x^2}.$

On separating we can write,

$I=\frac{5}{6}\int\frac{2+6x}{1+2x+3x^2}-\frac{11}{3}\int\frac{dx}{1+2x+3x^2}.$

Put 1+2x+3x^2=t.$On differentiating (2+6x)dx=dt.$1+2x+3x^2=3(x^2+\frac{2}{3}+\frac{1}{3})$.$\qquad=3\big(x+\frac{1}{3}\big)^2+\big(\frac{\sqrt 2}{3}\big)^2$On substituting for t and dt,$I=\frac{5}{6}\int\frac{dt}{t}-\frac{11}{3}\int\frac{dx}{3(x+1/3)^2+\big(\frac{\sqrt 2}{3}}\big)^2$.$\;\;\;=\frac{5}{6}\int\frac{dt}{t}-\frac{11}{9}\int\frac{dx}{(x+1/3)^2+\big(\frac{\sqrt 2}{3}}\big)^2$. On integrating we get,$\;\;\;=\frac{5}{6}log|t|-\frac{11}{9}\frac{3}{\sqrt 2}\tan^{-1}\big(\frac{x+1/3}{\sqrt 2/3}\big).$Substituting for t we get,$\;\;\;=\frac{5}{6}log(1+2x+3x^2)-\frac{11}{3\sqrt 2}\tan^{-1}\frac{(3x+1)}{\sqrt 2}\$.