What is the integral of (sqrt(x))/(1+x^{3/2)} ?

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

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integral of (sqrt(x))/(1+x^{3/2)}

\int \frac{x}{1 + x ^{\frac{3}{2}}} d x

\frac{2}{3} ln 1 + x^{\frac{3}{2}} + C

Solution steps

\int \frac{x}{1 + x ^{\frac{3}{2}}} d x

Apply u-substitution

: \int \frac{2}{3 u} d u

= \int \frac{2}{3 u} d u

Take the constant out:

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

= \frac{2}{3} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{2}{3} ln ∣ u ∣

Substitute back

u = 1 + x^{\frac{3}{2}}

= \frac{2}{3} ln 1 + x^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{3} ln 1 + x^{\frac{3}{2}} + C

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