What is the integral of x^3*sqrt(x^2+1) ?

The integral of x^3*sqrt(x^2+1) is 1/15 (3x^2-2)(x^2+1)^{3/2}+C

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integral of x^3*sqrt(x^2+1)

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

\frac{1}{15} (3 x^{2} - 2) (x^{2} + 1)^{\frac{3}{2}} + C

Solution steps

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

Apply u-substitution

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

= \int \frac{( u - 1 ) u}{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) u d u

Expand

(u - 1) u : u^{\frac{3}{2}} - u

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int u^{\frac{3}{2}} d u - \int u d u)

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

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

= \frac{1}{2} (\frac{2}{5} u^{\frac{5}{2}} - \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = x^{2} + 1

= \frac{1}{2} (\frac{2}{5} (x^{2} + 1)^{\frac{5}{2}} - \frac{2}{3} (x^{2} + 1)^{\frac{3}{2}})

Simplify

\frac{1}{2} (\frac{2}{5} (x^{2} + 1)^{\frac{5}{2}} - \frac{2}{3} (x^{2} + 1)^{\frac{3}{2}}) : \frac{1}{15} (3 x^{2} - 2) (x^{2} + 1)^{\frac{3}{2}}

= \frac{1}{15} (3 x^{2} - 2) (x^{2} + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{15} (3 x^{2} - 2) (x^{2} + 1)^{\frac{3}{2}} + C

x^{3} - 8 x^{2} + 5 x + 8

\frac{1}{2 s ^{2}}

\frac{d}{d x} (\frac{1}{3} x^{6} (x^{2} - 3))

\frac{d y}{d x} + x y = e^{x} y

\frac{d}{d x} (\frac{( x ^{4} - 4 x + 3 )}{x + 1})

What is the integral of x^3*sqrt(x^2+1) ?
The integral of x^3*sqrt(x^2+1) is 1/15 (3x^2-2)(x^2+1)^{3/2}+C