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

The integral of (x^2)/((3x+1)^3) is 1/27 ln|3x+1|+2/(27(3x+1))-1/(54(3x+1)^2)+C

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

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

\frac{1}{27} ln ∣ 3 x + 1 ∣ + \frac{2}{27 ( 3 x + 1 )} - \frac{1}{54 ( 3 x + 1 ) ^{2}} + C

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{9 ( 3 x + 1 )} d x = \frac{1}{27} ln ∣ 3 x + 1 ∣

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

\int \frac{1}{9 ( 3 x + 1 ) ^{3}} d x = - \frac{1}{54 ( 3 x + 1 ) ^{2}}

= \frac{1}{27} ln ∣ 3 x + 1 ∣ - (- \frac{2}{27 ( 3 x + 1 )}) - \frac{1}{54 ( 3 x + 1 ) ^{2}}

Simplify

= \frac{1}{27} ln ∣ 3 x + 1 ∣ + \frac{2}{27 ( 3 x + 1 )} - \frac{1}{54 ( 3 x + 1 ) ^{2}}

Add a constant to the solution

= \frac{1}{27} ln ∣ 3 x + 1 ∣ + \frac{2}{27 ( 3 x + 1 )} - \frac{1}{54 ( 3 x + 1 ) ^{2}} + C

\frac{d}{d x} ((5 x)^{\frac{1}{2}})

\frac{d}{d x} (x cos (π x) + 1)

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = 0

x \to 4 + lim (\frac{2}{x ^{2} - 16})

x \to \infty lim (x^{3 x})

What is the integral of (x^2)/((3x+1)^3) ?
The integral of (x^2)/((3x+1)^3) is 1/27 ln|3x+1|+2/(27(3x+1))-1/(54(3x+1)^2)+C