DofotheroU
2021-11-07
Answered

Evaluate the integral.

$\int \frac{2{x}^{2}+7x-3}{x-2}dx$

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unessodopunsep

Answered 2021-11-08
Author has **105** answers

Given

The integral is$\int \frac{2{x}^{2}+7x-3}{x-2}dx$

solution

$\int \frac{2{x}^{2}+7x-3}{x-2}dx=\int \frac{2{(x-2)}^{2}+15(x-2)+19}{x-2}dx$

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

$\int \frac{2{x}^{2}+7x-3}{x-2}dx={(x-2)}^{2}+15(x-2)+19\mathrm{ln}(x-2)+C$

$={x}^{2}-4x+4+15x-30+19\mathrm{ln}(x-2)+C$

$={x}^{2}+11x-26+19\mathrm{ln}(x-2)+C$

$\int \frac{2{x}^{2}+7x-3}{x-2}dx={x}^{2}+11x+19\mathrm{ln}(x-2)+{C}^{\prime}\text{}\text{}\text{}({C}^{\prime}=C-26)$

The integral is

solution

let u=x-2; du=dx

put u=x-2

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