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

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

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

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

- \frac{5}{x + 1} + \frac{3}{2 ( x + 1 ) ^{2}} - 2 ln ∣ x ∣ + 2 ln ∣ x + 1 ∣ - \frac{1}{( x + 1 ) ^{2}} + C

Solution steps

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

Expand

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

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

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

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

= 3 (- \frac{1}{x + 1} + \frac{1}{2 ( x + 1 ) ^{2}}) - (2 ln ∣ x ∣ - 2 ln ∣ x + 1 ∣ + \frac{2}{x + 1} + \frac{1}{( x + 1 ) ^{2}})

Simplify

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

= - \frac{5}{x + 1} + \frac{3}{2 ( x + 1 ) ^{2}} - 2 ln ∣ x ∣ + 2 ln ∣ x + 1 ∣ - \frac{1}{( x + 1 ) ^{2}}

Add a constant to the solution

= - \frac{5}{x + 1} + \frac{3}{2 ( x + 1 ) ^{2}} - 2 ln ∣ x ∣ + 2 ln ∣ x + 1 ∣ - \frac{1}{( x + 1 ) ^{2}} + C

\int (5 x^{4} + 64) arctan (\frac{x}{2}) d x

f (x) = 5 - x

lo g_{10} (\frac{( x + 1 ) ( x - 8 )}{( x - 1 ) ( x + 8 )})

n = 1 \sum \infty \frac{1}{5 ^{n + 1}}

\int 3^{x} + \frac{2}{x} d x

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