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

The integral of 1/((x+3)sqrt(x+1)) is sqrt(2)arctan(sqrt(1/2 (x+1)))+C

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

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

2 arctan (\frac{1}{2} (x + 1)) + C

Solution steps

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

Apply u-substitution

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

= \int \frac{1}{u u - 2} d u

Apply u-substitution

: \int \frac{2}{v ^{2} + 2} d v

= \int \frac{2}{v ^{2} + 2} 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} + 2} d v

Apply Integral Substitution

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

Use the common integral:

\int \frac{1}{w ^{2} + 1} d w = arctan (w)

= 2 \cdot \frac{1}{2} arctan (w)

Substitute back

= 2 \cdot \frac{1}{2} arctan (\frac{x + 3 - 2}{2})

Simplify

2 \cdot \frac{1}{2} arctan (\frac{x + 3 - 2}{2}) : 2 arctan (\frac{1}{2} (x + 1))

= 2 arctan (\frac{1}{2} (x + 1))

Add a constant to the solution

= 2 arctan (\frac{1}{2} (x + 1)) + C

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What is the integral of 1/((x+3)sqrt(x+1)) ?
The integral of 1/((x+3)sqrt(x+1)) is sqrt(2)arctan(sqrt(1/2 (x+1)))+C