What is the integral of (x^2)/(x^3+x) ?

The integral of (x^2)/(x^3+x) is 1/2 ln|x^2+1|+C

Popular >

integral of (x^2)/(x^3+x)

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

\frac{1}{2} ln x^{2} + 1 + C

Solution steps

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

Factor

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

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

Cancel the common factor:

x

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} + 1

= \frac{1}{2} ln x^{2} + 1

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 1 + C

\int θ tan (2) θ^{2} d θ

\int \frac{- x}{x ^{2} - x + 1} d x

\int \frac{x}{25 x ^{4} + 9} d x

\int \frac{1}{5 + 4 sin ( 2 x )} d x

\int \frac{4 x - 3}{( x - 2 ) ( x ^{2} + 1 )} d x

What is the integral of (x^2)/(x^3+x) ?
The integral of (x^2)/(x^3+x) is 1/2 ln|x^2+1|+C