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

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

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

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

2 arctan (x) + C

Solution steps

\int \frac{1}{x ( x + 1 )} 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 = x

= 2 arctan (x)

Add a constant to the solution

= 2 arctan (x) + C

2 x e^{x^{2}}

\int x^{3} e^{x y}

\int_{6}^{8} \frac{44}{( x - 6 ) ^{3}} d x

\frac{d}{d x} (- \frac{10}{x ^{3}})

\int_{0}^{\frac{π}{4}} sec^{3} (x) d x

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