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

The integral of ((2x+1))/((x+2)^3(x-1)^3) is -1/(27(x+2))-1/(18(x+2)^2)+1/(27(x-1))-1/(18(x-1)^2)+C

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

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

- \frac{1}{27 ( x + 2 )} - \frac{1}{18 ( x + 2 ) ^{2}} + \frac{1}{27 ( x - 1 )} - \frac{1}{18 ( x - 1 ) ^{2}} + C

Solution steps

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

Take the partial fraction of

\frac{2 x + 1}{( x + 2 ) ^{3} ( x - 1 ) ^{3}} : \frac{1}{27 ( x + 2 ) ^{2}} + \frac{1}{9 ( x + 2 ) ^{3}} - \frac{1}{27 ( x - 1 ) ^{2}} + \frac{1}{9 ( x - 1 ) ^{3}}

= \int \frac{1}{27 ( x + 2 ) ^{2}} + \frac{1}{9 ( x + 2 ) ^{3}} - \frac{1}{27 ( x - 1 ) ^{2}} + \frac{1}{9 ( x - 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{1}{27 ( x + 2 ) ^{2}} d x + \int \frac{1}{9 ( x + 2 ) ^{3}} d x - \int \frac{1}{27 ( x - 1 ) ^{2}} d x + \int \frac{1}{9 ( x - 1 ) ^{3}} d x

\int \frac{1}{27 ( x + 2 ) ^{2}} d x = - \frac{1}{27 ( x + 2 )}

\int \frac{1}{9 ( x + 2 ) ^{3}} d x = - \frac{1}{18 ( x + 2 ) ^{2}}

\int \frac{1}{27 ( x - 1 ) ^{2}} d x = - \frac{1}{27 ( x - 1 )}

\int \frac{1}{9 ( x - 1 ) ^{3}} d x = - \frac{1}{18 ( x - 1 ) ^{2}}

= - \frac{1}{27 ( x + 2 )} - \frac{1}{18 ( x + 2 ) ^{2}} - (- \frac{1}{27 ( x - 1 )}) - \frac{1}{18 ( x - 1 ) ^{2}}

Simplify

= - \frac{1}{27 ( x + 2 )} - \frac{1}{18 ( x + 2 ) ^{2}} + \frac{1}{27 ( x - 1 )} - \frac{1}{18 ( x - 1 ) ^{2}}

Add a constant to the solution

= - \frac{1}{27 ( x + 2 )} - \frac{1}{18 ( x + 2 ) ^{2}} + \frac{1}{27 ( x - 1 )} - \frac{1}{18 ( x - 1 ) ^{2}} + C

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What is the integral of ((2x+1))/((x+2)^3(x-1)^3) ?
The integral of ((2x+1))/((x+2)^3(x-1)^3) is -1/(27(x+2))-1/(18(x+2)^2)+1/(27(x-1))-1/(18(x-1)^2)+C