How do you find $\int \frac{x^{2} + 1}{{(x + 1)}^{3}} d x$ $int (x^2+1)/(x+1)^3dx$ using partial fractions?

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Answer 1 · 2018-05-31T17:16:10

$x - \frac{2}{x + 1} - 2 ln | x + 1 | + C$ $x-2/(x+1)-2ln|x+1|+C$

Note that

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

Answer 2 · 2018-05-31T19:08:55

$ln (x + 1) + \frac{2}{x + 1} - \frac{1}{{(x + 1)}^{2}} + C$ $ln(x+1)+2/(x+1)-1/(x+1)^2+C$

Due to $x^{2} + 1 = x^{2} + 2 x + 1 - 2 x - 2 + 2$ $x^2+1=x^2+2x+1-2x-2+2$ or ${(x + 1)}^{2} - 2 \cdot (x + 1) + 2$ $(x+1)^2-2*(x+1)+2$

$\int \frac{x^{2} + 1}{{(x + 1)}^{3}} \cdot d x$ $int (x^2+1)/(x+1)^3*dx$

= $\int \frac{d x}{x + 1} - 2 \int \frac{d x}{{(x + 1)}^{2}} + 2 \int \frac{d x}{{(x + 1)}^{2}}$ $int dx/(x+1)-2int dx/(x+1)^2+2int dx/(x+1)^2$

= $ln (x + 1) + \frac{2}{x + 1} - \frac{1}{{(x + 1)}^{2}} + C$ $ln(x+1)+2/(x+1)-1/(x+1)^2+C$

How do you find int (x^2+1)/(x+1)^3dx∫x2+1(x+1)3dxint (x^2+1)/(x+1)^3dx using partial fractions?