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

The integral of (3x^2+1)/((x+2)^3) is 3(ln|x+2|+4/(x+2)-2/((x+2)^2))-1/(2(x+2)^2)+C

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

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

3 (ln ∣ x + 2 ∣ + \frac{4}{x + 2} - \frac{2}{( x + 2 ) ^{2}}) - \frac{1}{2 ( x + 2 ) ^{2}} + C

Solution steps

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

Expand

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

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

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

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

= 3 (ln ∣ x + 2 ∣ + \frac{4}{x + 2} - \frac{2}{( x + 2 ) ^{2}}) - \frac{1}{2 ( x + 2 ) ^{2}}

Add a constant to the solution

= 3 (ln ∣ x + 2 ∣ + \frac{4}{x + 2} - \frac{2}{( x + 2 ) ^{2}}) - \frac{1}{2 ( x + 2 ) ^{2}} + C

What is the integral of (3x^2+1)/((x+2)^3) ?
The integral of (3x^2+1)/((x+2)^3) is 3(ln|x+2|+4/(x+2)-2/((x+2)^2))-1/(2(x+2)^2)+C