What is the integral of x^3sqrt(x^2+42) ?

The integral of x^3sqrt(x^2+42) is 1/5 (x^2+42)^{5/2}-14(x^2+42)^{3/2}+C

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integral of x^3sqrt(x^2+42)

\int x^{3} x^{2} + 42 d x

\frac{1}{5} (x^{2} + 42)^{\frac{5}{2}} - 14 (x^{2} + 42)^{\frac{3}{2}} + C

Solution steps

\int x^{3} x^{2} + 42 d x

Apply u-substitution

: \int (u^{2} - 42) u^{2} d u

= \int (u^{2} - 42) u^{2} d u

Expand

(u^{2} - 42) u^{2} : u^{4} - 42 u^{2}

= \int u^{4} - 42 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 42 u^{2} d u

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

\int 42 u^{2} d u = 14 u^{3}

= \frac{u ^{5}}{5} - 14 u^{3}

Substitute back

u = x^{2} + 42

= \frac{( x ^{2} + 42 ) ^{5}}{5} - 14 (x^{2} + 42)^{3}

Simplify

\frac{( x ^{2} + 42 ) ^{5}}{5} - 14 (x^{2} + 42)^{3} : \frac{1}{5} (x^{2} + 42)^{\frac{5}{2}} - 14 (x^{2} + 42)^{\frac{3}{2}}

= \frac{1}{5} (x^{2} + 42)^{\frac{5}{2}} - 14 (x^{2} + 42)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{5} (x^{2} + 42)^{\frac{5}{2}} - 14 (x^{2} + 42)^{\frac{3}{2}} + C

x \to 1 lim (\frac{x ^{2} + x}{x - 1})

6 t \frac{d y}{d t} + y = t^{5}

\frac{d}{d x} (1 + cos^{2} (x))

\int (x - 2) e^{x} d x

y = 2 arcsin (3) x^{5}

What is the integral of x^3sqrt(x^2+42) ?
The integral of x^3sqrt(x^2+42) is 1/5 (x^2+42)^{5/2}-14(x^2+42)^{3/2}+C