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

The integral of x^3sqrt(x^2+36) is 1/5 (x^2+36)^{5/2}-12(x^2+36)^{3/2}+C

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

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

\frac{1}{5} (x^{2} + 36)^{\frac{5}{2}} - 12 (x^{2} + 36)^{\frac{3}{2}} + C

Solution steps

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

Apply u-substitution

: \int (u^{2} - 36) u^{2} d u

= \int (u^{2} - 36) u^{2} d u

Expand

(u^{2} - 36) u^{2} : u^{4} - 36 u^{2}

= \int u^{4} - 36 u^{2} 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

= \int u^{4} d u - \int 36 u^{2} d u

\int u^{4} d u = \frac{u ^{5}}{5}

\int 36 u^{2} d u = 12 u^{3}

= \frac{u ^{5}}{5} - 12 u^{3}

Substitute back

u = x^{2} + 36

= \frac{( x ^{2} + 36 ) ^{5}}{5} - 12 (x^{2} + 36)^{3}

Simplify

\frac{( x ^{2} + 36 ) ^{5}}{5} - 12 (x^{2} + 36)^{3} : \frac{1}{5} (x^{2} + 36)^{\frac{5}{2}} - 12 (x^{2} + 36)^{\frac{3}{2}}

= \frac{1}{5} (x^{2} + 36)^{\frac{5}{2}} - 12 (x^{2} + 36)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{5} (x^{2} + 36)^{\frac{5}{2}} - 12 (x^{2} + 36)^{\frac{3}{2}} + C

x \to 0 lim (\frac{( 2 x ^{3} )}{2 x})

\int e^{2 x} \cdot 2 d x

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

y^{^{''}} + 4 y^{^{'}} + 5 y = 15 x + 3 e^{- x}

y^{^{''}} + 8 y^{^{'}} + 7 y = - 8 - x + 24 x^{2} + 7 x^{3}

What is the integral of x^3sqrt(x^2+36) ?
The integral of x^3sqrt(x^2+36) is 1/5 (x^2+36)^{5/2}-12(x^2+36)^{3/2}+C