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

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

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

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

- ln ∣ 3 x + 5 ∣ + ln ∣ 3 x + 3 ∣ + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Complete the square

u (u - 2) : (u - 1)^{2} - 1

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

Apply u-substitution

: \int \frac{1}{v ^{2} - 1} d v

= 2 \cdot \int \frac{1}{v ^{2} - 1} d v

Factor

v^{2} - 1 : - (- v^{2} + 1)

= 2 \cdot \int \frac{1}{- ( - v ^{2} + 1 )} d v

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= 2 (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

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

Simplify

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

= - ln ∣ 3 x + 5 ∣ + ln ∣ 3 x + 3 ∣

Add a constant to the solution

= - ln ∣ 3 x + 5 ∣ + ln ∣ 3 x + 3 ∣ + C

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