What is the integral of (180x^2)/((3x+1)(3x+5)(x+2)) ?

The integral of (180x^2)/((3x+1)(3x+5)(x+2)) is ln|3x+1|-125ln|3x+5|+144ln|x+2|+C

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integral of (180x^2)/((3x+1)(3x+5)(x+2))

\int \frac{180 x ^{2}}{( 3 x + 1 ) ( 3 x + 5 ) ( x + 2 )} d x

ln ∣ 3 x + 1 ∣ - 125 ln ∣ 3 x + 5 ∣ + 144 ln ∣ x + 2 ∣ + C

Solution steps

\int \frac{180 x ^{2}}{( 3 x + 1 ) ( 3 x + 5 ) ( x + 2 )} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 180 \cdot \int \frac{x ^{2}}{( 3 x + 1 ) ( 3 x + 5 ) ( x + 2 )} d x

Take the partial fraction of

\frac{x ^{2}}{( 3 x + 1 ) ( 3 x + 5 ) ( x + 2 )} : \frac{1}{60 ( 3 x + 1 )} - \frac{25}{12 ( 3 x + 5 )} + \frac{4}{5 ( x + 2 )}

= 180 \cdot \int \frac{1}{60 ( 3 x + 1 )} - \frac{25}{12 ( 3 x + 5 )} + \frac{4}{5 ( x + 2 )} 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

= 180 (\int \frac{1}{60 ( 3 x + 1 )} d x - \int \frac{25}{12 ( 3 x + 5 )} d x + \int \frac{4}{5 ( x + 2 )} d x)

\int \frac{1}{60 ( 3 x + 1 )} d x = \frac{1}{180} ln ∣ 3 x + 1 ∣

\int \frac{25}{12 ( 3 x + 5 )} d x = \frac{25}{36} ln ∣ 3 x + 5 ∣

\int \frac{4}{5 ( x + 2 )} d x = \frac{4}{5} ln ∣ x + 2 ∣

= 180 (\frac{1}{180} ln ∣ 3 x + 1 ∣ - \frac{25}{36} ln ∣ 3 x + 5 ∣ + \frac{4}{5} ln ∣ x + 2 ∣)

Simplify

180 (\frac{1}{180} ln ∣ 3 x + 1 ∣ - \frac{25}{36} ln ∣ 3 x + 5 ∣ + \frac{4}{5} ln ∣ x + 2 ∣) : ln ∣ 3 x + 1 ∣ - 125 ln ∣ 3 x + 5 ∣ + 144 ln ∣ x + 2 ∣

= ln ∣ 3 x + 1 ∣ - 125 ln ∣ 3 x + 5 ∣ + 144 ln ∣ x + 2 ∣

Add a constant to the solution

= ln ∣ 3 x + 1 ∣ - 125 ln ∣ 3 x + 5 ∣ + 144 ln ∣ x + 2 ∣ + C

f (x) = cos (ln (x))

\int_{0}^{4} \frac{1}{x ^{2}} d x

\frac{\partial}{\partial x} (ln (x^{2}))

\int - 2 cos (t) d t

What is the integral of (180x^2)/((3x+1)(3x+5)(x+2)) ?
The integral of (180x^2)/((3x+1)(3x+5)(x+2)) is ln|3x+1|-125ln|3x+5|+144ln|x+2|+C