What is the integral of x^3sqrt(x^2+22) ?

The integral of x^3sqrt(x^2+22) is 1/5 (x^2+22)^{5/2}-22/3 (x^2+22)^{3/2}+C

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integral of x^3sqrt(x^2+22)

\int x^{3} x^{2} + 22 d x

\frac{1}{5} (x^{2} + 22)^{\frac{5}{2}} - \frac{22}{3} (x^{2} + 22)^{\frac{3}{2}} + C

Solution steps

\int x^{3} x^{2} + 22 d x

Apply u-substitution

: \int (u^{2} - 22) u^{2} d u

= \int (u^{2} - 22) u^{2} d u

Expand

(u^{2} - 22) u^{2} : u^{4} - 22 u^{2}

= \int u^{4} - 22 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 22 u^{2} d u

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

\int 22 u^{2} d u = \frac{22 u ^{3}}{3}

= \frac{u ^{5}}{5} - \frac{22 u ^{3}}{3}

Substitute back

u = x^{2} + 22

= \frac{( x ^{2} + 22 ) ^{5}}{5} - \frac{22 ( x ^{2} + 22 ) ^{3}}{3}

Simplify

\frac{( x ^{2} + 22 ) ^{5}}{5} - \frac{22 ( x ^{2} + 22 ) ^{3}}{3} : \frac{1}{5} (x^{2} + 22)^{\frac{5}{2}} - \frac{22}{3} (x^{2} + 22)^{\frac{3}{2}}

= \frac{1}{5} (x^{2} + 22)^{\frac{5}{2}} - \frac{22}{3} (x^{2} + 22)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{5} (x^{2} + 22)^{\frac{5}{2}} - \frac{22}{3} (x^{2} + 22)^{\frac{3}{2}} + C

y^{^{''}} + 4 π^{2} y = cos^{3} (2 π x), y (10) = 2

y^{2} = x + 1, x = 3

x \to 2 - lim (\frac{x - 3}{x + 2})

\int (4 t^{3}) d t

f (x) = \frac{1}{x} + \frac{1}{1 - x}

What is the integral of x^3sqrt(x^2+22) ?
The integral of x^3sqrt(x^2+22) is 1/5 (x^2+22)^{5/2}-22/3 (x^2+22)^{3/2}+C