 # Evaluate the integral. \int \frac{2x^{2}+7x-3}{x-2}dx DofotheroU 2021-11-07 Answered
Evaluate the integral.
$\int \frac{2{x}^{2}+7x-3}{x-2}dx$
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Given
The integral is $\int \frac{2{x}^{2}+7x-3}{x-2}dx$
solution
$\int \frac{2{x}^{2}+7x-3}{x-2}dx=\int \frac{2{\left(x-2\right)}^{2}+15\left(x-2\right)+19}{x-2}dx$
let u=x-2; du=dx
$\int \frac{2{x}^{2}+7x-3}{x-2}dx=\int \frac{2{u}^{2}+15u+19}{u}du$
$=\int \frac{2{u}^{2}}{u}du+\int \frac{15u}{u}du+\int \frac{19}{u}du$
$=2\int udu+15\int 1du+19\int \frac{1}{u}du$
$=2\cdot \frac{{u}^{2}}{2}+15u+19\mathrm{ln}u+C$
put u=x-2
$\int \frac{2{x}^{2}+7x-3}{x-2}dx={\left(x-2\right)}^{2}+15\left(x-2\right)+19\mathrm{ln}\left(x-2\right)+C$
$={x}^{2}-4x+4+15x-30+19\mathrm{ln}\left(x-2\right)+C$
$={x}^{2}+11x-26+19\mathrm{ln}\left(x-2\right)+C$