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

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

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integral of 1/((2+x)sqrt(1+x))

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

2 arctan (1 + x) + C

Solution steps

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

Apply u-substitution

: \int \frac{2}{u ^{2} + 1} d u

= \int \frac{2}{u ^{2} + 1} d u

Take the constant out:

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

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

Use the common integral:

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

= 2 arctan (u)

Substitute back

u = 1 + x

= 2 arctan (1 + x)

Add a constant to the solution

= 2 arctan (1 + x) + C

f (x) = x^{4} - 2 x^{2} + 1, at x = 2

\int_{0}^{π} sin (x) sin (3 x) d x

\frac{x ^{2}}{x ^{2} + 3}

y^{^{'}} = x^{2} y - y + x^{2} - 1

\int e [2 x] d x

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