What is the integral of x^3sqrt(x^2+35) ?

The integral of x^3sqrt(x^2+35) is 1/5 (x^2+35)^{5/2}-35/3 (x^2+35)^{3/2}+C

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integral of x^3sqrt(x^2+35)

\int x^{3} x^{2} + 35 d x

\frac{1}{5} (x^{2} + 35)^{\frac{5}{2}} - \frac{35}{3} (x^{2} + 35)^{\frac{3}{2}} + C

Solution steps

\int x^{3} x^{2} + 35 d x

Apply u-substitution

: \int (u^{2} - 35) u^{2} d u

= \int (u^{2} - 35) u^{2} d u

Expand

(u^{2} - 35) u^{2} : u^{4} - 35 u^{2}

= \int u^{4} - 35 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 35 u^{2} d u

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

\int 35 u^{2} d u = \frac{35 u ^{3}}{3}

= \frac{u ^{5}}{5} - \frac{35 u ^{3}}{3}

Substitute back

u = x^{2} + 35

= \frac{( x ^{2} + 35 ) ^{5}}{5} - \frac{35 ( x ^{2} + 35 ) ^{3}}{3}

Simplify

\frac{( x ^{2} + 35 ) ^{5}}{5} - \frac{35 ( x ^{2} + 35 ) ^{3}}{3} : \frac{1}{5} (x^{2} + 35)^{\frac{5}{2}} - \frac{35}{3} (x^{2} + 35)^{\frac{3}{2}}

= \frac{1}{5} (x^{2} + 35)^{\frac{5}{2}} - \frac{35}{3} (x^{2} + 35)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{5} (x^{2} + 35)^{\frac{5}{2}} - \frac{35}{3} (x^{2} + 35)^{\frac{3}{2}} + C

e^{(x^{3})}

x^{2}, x = 0, x = 1

tan^{2} (sin^{4} (t))

\frac{3}{s ( s ^{2} + 8 s + 18 )}

x \to 2 - lim (x^{2} - 4)

What is the integral of x^3sqrt(x^2+35) ?
The integral of x^3sqrt(x^2+35) is 1/5 (x^2+35)^{5/2}-35/3 (x^2+35)^{3/2}+C