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

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

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

integral

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

\frac{1}{3} (x^{2} + 1)^{\frac{3}{2}} + C

Solution steps

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

Apply u-substitution

: \int \frac{( u + 1 ) ^{\frac{1}{2}}}{2} d u

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

Apply u-substitution

: \int v^{\frac{1}{2}} d v

= \frac{1}{2} \cdot \int v^{\frac{1}{2}} d v

Apply the Power Rule

: \frac{2}{3} v^{\frac{3}{2}}

= \frac{1}{2} \cdot \frac{2}{3} v^{\frac{3}{2}}

Substitute back

= \frac{1}{2} \cdot \frac{2}{3} (x^{2} + 1)^{\frac{3}{2}}

Simplify

\frac{1}{2} \cdot \frac{2}{3} (x^{2} + 1)^{\frac{3}{2}} : \frac{1}{3} (x^{2} + 1)^{\frac{3}{2}}

= \frac{1}{3} (x^{2} + 1)^{\frac{3}{2}}

Add a constant to the solution

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

ln (a^{2} - x^{2})

(x^{3} - 1)^{5}

\frac{x ^{2}}{x + 4}

y^{2} \cdot 2 (x^{2} - y^{2})^{\frac{1}{2}}

\frac{2 ln ( x )}{x}

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