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

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

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

\int \frac{x}{x ^{2} + x} d x

2 arctan (x) + C

Solution steps

\int \frac{x}{x ^{2} + x} d x

Simplify

\frac{x}{x ^{2} + x} : \frac{1}{x ( x + 1 )}

= \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

\frac{d}{d x} (x^{2} \cdot e^{- x^{2}})

f (x) = - x^{2} - 2 x - 2, at x = - 3

\int \frac{I n ( x )}{x ^{2}} d x

\int \frac{x ^{3}}{x ^{2} + 5} d x

\int (2 x + 4) d x

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