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 \cdot 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

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

x \to \infty lim (- (e^{x}))

\int \frac{7}{x ( x ^{4} + 1 )} d x

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (0) = - 3, y^{^{'}} (0) = 14

f (x) = \frac{4}{x}, at x = \frac{1}{16}

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