What is the integral of (x^5+2x^3-3x)/((x^2+1)^3) ?

The integral of (x^5+2x^3-3x)/((x^2+1)^3) is 1/2 ln|x^2+1|+1/((x^2+1)^2)+C

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integral of (x^5+2x^3-3x)/((x^2+1)^3)

\int \frac{x ^{5} + 2 x ^{3} - 3 x}{( x ^{2} + 1 ) ^{3}} d x

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

Solution steps

\int \frac{x ^{5} + 2 x ^{3} - 3 x}{( x ^{2} + 1 ) ^{3}} d x

Take the partial fraction of

\frac{x ^{5} + 2 x ^{3} - 3 x}{( x ^{2} + 1 ) ^{3}} : \frac{x}{x ^{2} + 1} - \frac{4 x}{( x ^{2} + 1 ) ^{3}}

= \int \frac{x}{x ^{2} + 1} - \frac{4 x}{( x ^{2} + 1 ) ^{3}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{x}{x ^{2} + 1} d x - \int \frac{4 x}{( x ^{2} + 1 ) ^{3}} d x

\int \frac{x}{x ^{2} + 1} d x = \frac{1}{2} ln x^{2} + 1

\int \frac{4 x}{( x ^{2} + 1 ) ^{3}} d x = - \frac{1}{( x ^{2} + 1 ) ^{2}}

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

Simplify

= \frac{1}{2} ln x^{2} + 1 + \frac{1}{( x ^{2} + 1 ) ^{2}}

Add a constant to the solution

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

\int 2 e^{y^{2}} d y

e^{2 x^{x}}

\int_{1}^{4} (x + x) d x

\int \frac{( x ^{4} + 81 )}{x ( x ^{2} + 9 ) ^{2}} d x

\frac{1}{s ^{2} - 16}

What is the integral of (x^5+2x^3-3x)/((x^2+1)^3) ?
The integral of (x^5+2x^3-3x)/((x^2+1)^3) is 1/2 ln|x^2+1|+1/((x^2+1)^2)+C