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

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

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

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

2 arctan (x) + C

Solution steps

\int \frac{1}{( 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 = x

= 2 arctan (x)

Add a constant to the solution

= 2 arctan (x) + C

\frac{\partial}{\partial t} (\frac{1}{t})

\int cos (I n (x)) d x

\frac{d}{d x} (\frac{1}{( 1 - 2 x ) ^{2}})

\frac{d}{d x} (\frac{x ^{2} - 64}{x ^{2} + 64})

x \to 0 lim (\frac{\frac{4}{1 + x} - 4}{x})

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