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

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

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integral of sqrt(x^2+1)x

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

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

Solution steps

\int x^{2} + 1 x 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

x \to 0 lim (sin (x) \cdot ln (4 x))

\frac{\partial}{\partial x} (2 x + y - 5)

\frac{d}{d x} (arccos (\frac{y}{x}))

y^{^{'}} = y \cdot tan (x) + sin (x)

\frac{\partial}{\partial x} (x ln (x^{2} y + y^{2}))

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