Question

evaluate Integration of 3x-2/x^2-3x+2 dx​

Answer 1

Correct Question:

Integrate:

$\displaystyle \rm \hookrightarrow I = \int \dfrac{2x + 3}{ \sqrt{ {x}^{2} + 3x - 2 } } \: dx$

Solution:

Given Integral:

$\displaystyle \rm \longrightarrow I = \int \dfrac{2x + 3}{ \sqrt{ {x}^{2} + 3x - 2 } } \: dx$

Let us assume that:

$\rm \longrightarrow u = {x}^{2} + 3x - 2$

$\rm \longrightarrow \dfrac{du}{dx} = 2x + 3$

$\rm \longrightarrow du = (2x + 3) \: dx$

Therefore, the integral changes to:

$\displaystyle \rm \longrightarrow I = \int \dfrac{du}{ \sqrt{u} }$

$\displaystyle \rm \longrightarrow I = \dfrac{ {u}^{ \frac{ - 1}{2} + 1 } }{ \frac{ - 1}{2} + 1} + C$

$\displaystyle \rm \longrightarrow I = \dfrac{ {u}^{ \frac{ 1}{2}} }{ \frac{1}{2} } + C$

$\displaystyle \rm \longrightarrow I =2 \sqrt{u} + C$

Substituting back u = x² + 3x - 2, we get:

$\displaystyle \rm \longrightarrow I =2 \sqrt{ {x}^{2} + 3x - 2 } + C$

Therefore:

$\displaystyle \rm \longrightarrow \int \frac{2x + 3}{ \sqrt{ {x}^{2} +3x - 2 } } \: dx =2 \sqrt{ {x}^{2} + 3x - 2 } + C$

★ Which is our required answer.

Additional Information:

$\boxed{\begin{array}{c|c}\bf f(x)&\bf\displaystyle\int\rm f(x)\:dx\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \rm k&\rm kx+C\\ \\ \rm sin(x)&\rm-cos(x)+C\\ \\ \rm cos(x)&\rm sin(x)+C\\ \\ \rm{sec}^{2}(x)&\rm tan(x)+C\\ \\ \rm{cosec}^{2}(x)&\rm-cot(x)+C\\ \\ \rm sec(x)\ tan(x)&\rm sec(x)+C\\ \\ \rm cosec(x)\ cot(x)&\rm-cosec(x)+C\\ \\ \rm tan(x)&\rm log(sec(x))+C\\ \\ \rm\dfrac{1}{x}&\rm log(x)+C\\ \\ \rm{e}^{x}&\rm{e}^{x}+C\\ \\ \rm x^{n},n\neq-1&\rm\dfrac{x^{n+1}}{n+1}+C\end{array}}$