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

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

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

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

\frac{1}{2} ln ∣ 3 x + 1 ∣ - \frac{1}{2} ln ∣ 3 x + 3 ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ( u + 2 )} d u

= \int \frac{1}{u ( u + 2 )} d u

Take the partial fraction of

\frac{1}{u ( u + 2 )} : \frac{1}{2 u} - \frac{1}{2 ( u + 2 )}

= \int \frac{1}{2 u} - \frac{1}{2 ( u + 2 )} d u

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}{2 u} d u - \int \frac{1}{2 ( u + 2 )} d u

\int \frac{1}{2 u} d u = \frac{1}{2} ln ∣ u ∣

\int \frac{1}{2 ( u + 2 )} d u = \frac{1}{2} ln ∣ u + 2 ∣

= \frac{1}{2} ln ∣ u ∣ - \frac{1}{2} ln ∣ u + 2 ∣

Substitute back

u = 3 x + 1

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

Simplify

\frac{1}{2} ln ∣ 3 x + 1 ∣ - \frac{1}{2} ln ∣ 3 x + 1 + 2 ∣ : \frac{1}{2} ln ∣ 3 x + 1 ∣ - \frac{1}{2} ln ∣ 3 x + 3 ∣

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

Add a constant to the solution

= \frac{1}{2} ln ∣ 3 x + 1 ∣ - \frac{1}{2} ln ∣ 3 x + 3 ∣ + C

\int \frac{x}{5 + x - x ^{2}} d x

\int x 2 x + 3 d x

\frac{\partial}{\partial x} (cos^{6} (x^{9} y^{5}))

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

y^{^{''}} + 2 y^{^{'}} + 2 y = 8 e^{2 t}, y (0) = 0, y^{^{'}} (0) = 1

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