What is the integral of 1/((x+1)sqrt(x+6)) ?

The integral of 1/((x+1)sqrt(x+6)) is -1/(sqrt(5))(ln|sqrt((x+6)/5)+1|-ln|sqrt((x+6)/5)-1|)+C

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integral of 1/((x+1)sqrt(x+6))

\int \frac{1}{( x + 1 ) x + 6} d x

- \frac{1}{5} (ln \frac{x + 6}{5} + 1 - ln \frac{x + 6}{5} - 1) + C

Solution steps

\int \frac{1}{( x + 1 ) x + 6} d x

Apply u-substitution

= \int \frac{1}{u u + 5} d u

Apply u-substitution

= \int \frac{2}{v ^{2} - 5} d v

Take the constant out:

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

= 2 \cdot \int \frac{1}{v ^{2} - 5} d v

Apply Integral Substitution

= 2 \cdot \int \frac{1}{5 ( w ^{2} - 1 )} d w

Take the constant out:

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

= 2 \cdot \frac{1}{5} \cdot \int \frac{1}{w ^{2} - 1} d w

Factor

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

= 2 \cdot \frac{1}{5} \cdot \int \frac{1}{- ( - w ^{2} + 1 )} d w

Take the constant out:

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

= 2 \cdot \frac{1}{5} (- \int \frac{1}{- w ^{2} + 1} d w)

Use the common integral:

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

= 2 \cdot \frac{1}{5} (- (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}))

Substitute back

= 2 \cdot \frac{1}{5} - \frac{ln \frac{x + 1 + 5}{5} + 1}{2} - \frac{ln \frac{x + 1 + 5}{5} - 1}{2}

Simplify

2 \cdot \frac{1}{5} - \frac{ln \frac{x + 1 + 5}{5} + 1}{2} - \frac{ln \frac{x + 1 + 5}{5} - 1}{2} : - \frac{1}{5} (ln \frac{x + 6}{5} + 1 - ln \frac{x + 6}{5} - 1)

= - \frac{1}{5} (ln \frac{x + 6}{5} + 1 - ln \frac{x + 6}{5} - 1)

Add a constant to the solution

= - \frac{1}{5} (ln \frac{x + 6}{5} + 1 - ln \frac{x + 6}{5} - 1) + C

What is the integral of 1/((x+1)sqrt(x+6)) ?
The integral of 1/((x+1)sqrt(x+6)) is -1/(sqrt(5))(ln|sqrt((x+6)/5)+1|-ln|sqrt((x+6)/5)-1|)+C