What is the integral of x^3sqrt(1+36x^2) ?

The integral of x^3sqrt(1+36x^2) is 1/9720 (54x^2-1)(36x^2+1)^{3/2}+C

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integral of x^3sqrt(1+36x^2)

\int x^{3} 1 + 36 x^{2} d x

\frac{1}{9720} (54 x^{2} - 1) (36 x^{2} + 1)^{\frac{3}{2}} + C

Solution steps

\int x^{3} 1 + 36 x^{2} d x

Apply u-substitution

: \int \frac{u ( u - 1 )}{2592} d u

= \int \frac{u ( u - 1 )}{2592} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2592} \cdot \int u (u - 1) d u

Expand

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

= \frac{1}{2592} \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}{2592} (\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}{2592} (\frac{2}{5} u^{\frac{5}{2}} - \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = 1 + 36 x^{2}

= \frac{1}{2592} (\frac{2}{5} (1 + 36 x^{2})^{\frac{5}{2}} - \frac{2}{3} (1 + 36 x^{2})^{\frac{3}{2}})

Simplify

\frac{1}{2592} (\frac{2}{5} (1 + 36 x^{2})^{\frac{5}{2}} - \frac{2}{3} (1 + 36 x^{2})^{\frac{3}{2}}) : \frac{1}{9720} (54 x^{2} - 1) (36 x^{2} + 1)^{\frac{3}{2}}

= \frac{1}{9720} (54 x^{2} - 1) (36 x^{2} + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{9720} (54 x^{2} - 1) (36 x^{2} + 1)^{\frac{3}{2}} + C

\int (e^{- \frac{x}{5}} + 5 x^{4}) d x

h (t) = (t + 4)^{\frac{2}{3}} (2 t^{2} - 4)^{3}

x \to 7 lim (sec (\frac{π x}{6}))

\frac{d y}{d x} = 7 x + y

x - x^{3}

What is the integral of x^3sqrt(1+36x^2) ?
The integral of x^3sqrt(1+36x^2) is 1/9720 (54x^2-1)(36x^2+1)^{3/2}+C