Question

Con procedimiento por favor

Answer 1

Respuesta:

$\frac{1}{x+1}+x+C$

Explicación paso a paso:

$\int \frac{x^2+2x}{\left(x+1\right)^2}dx$

$\frac{x^2+2x}{\left(x+1\right)^2}$

$=\frac{x^2+2x-\left(\left(x+1\right)^2\right)}{\left(x+1\right)^2}+1$

$=\frac{x^2+2x-\left(x+1\right)^2}{\left(x+1\right)^2}+1$

$=\frac{-1}{\left(x+1\right)^2}$

$=-\frac{1}{\left(x+1\right)^2}+1$

$=\int \:-\frac{1}{\left(x+1\right)^2}+1dx$

$\mathrm{Aplicar\:la\:regla\:de\:la\:suma}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx$

$-\int \frac{1}{\left(x+1\right)^2}dx+\int \:1dx$

$\int \frac{1}{\left(x+1\right)^2}dx$

$\mathrm{Sustituir:}\:u=x+1$

$\Rightarrow \:du=1dx$

$\Rightarrow \:dx=1du$

$=\int \frac{1}{u^2}\cdot \:1du$

$=\int \frac{1}{u^2}du$

$=\int \:u^{-2}du$

$=\frac{u^{-2+1}}{-2+1}$

$\mathrm{Sustituir\:en\:la\:ecuacion}\:u=x+1$

$=\frac{\left(x+1\right)^{-2+1}}{-2+1}$

$=-\frac{1}{x+1}$

$\int \:1dx$

$=1\cdot \:x$

$=x$

$=-\left(-\frac{1}{x+1}\right)+x$

$=\frac{1}{x+1}+x$

$=\frac{1}{x+1}+x+C$