What is the integral of x(sqrt(x^2+1)) ?

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

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integral of x(sqrt(x^2+1))

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

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

Solution steps

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

Apply u-substitution

: \int u^{2} d u

= \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= \frac{u ^{3}}{3}

Substitute back

u = x^{2} + 1

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

Simplify

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

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