What is the integral of x^3sqrt(x^2+41) ?

The integral of x^3sqrt(x^2+41) is 1/5 (x^2+41)^{5/2}-41/3 (x^2+41)^{3/2}+C

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integral of x^3sqrt(x^2+41)

\int x^{3} x^{2} + 41 d x

\frac{1}{5} (x^{2} + 41)^{\frac{5}{2}} - \frac{41}{3} (x^{2} + 41)^{\frac{3}{2}} + C

Solution steps

\int x^{3} x^{2} + 41 d x

Apply u-substitution

: \int (u^{2} - 41) u^{2} d u

= \int (u^{2} - 41) u^{2} d u

Expand

(u^{2} - 41) u^{2} : u^{4} - 41 u^{2}

= \int u^{4} - 41 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 41 u^{2} d u

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

\int 41 u^{2} d u = \frac{41 u ^{3}}{3}

= \frac{u ^{5}}{5} - \frac{41 u ^{3}}{3}

Substitute back

u = x^{2} + 41

= \frac{( x ^{2} + 41 ) ^{5}}{5} - \frac{41 ( x ^{2} + 41 ) ^{3}}{3}

Simplify

\frac{( x ^{2} + 41 ) ^{5}}{5} - \frac{41 ( x ^{2} + 41 ) ^{3}}{3} : \frac{1}{5} (x^{2} + 41)^{\frac{5}{2}} - \frac{41}{3} (x^{2} + 41)^{\frac{3}{2}}

= \frac{1}{5} (x^{2} + 41)^{\frac{5}{2}} - \frac{41}{3} (x^{2} + 41)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{5} (x^{2} + 41)^{\frac{5}{2}} - \frac{41}{3} (x^{2} + 41)^{\frac{3}{2}} + C

y = \frac{10}{( 10 - x ^{4} ) ^{\frac{3}{2}}}

\int \frac{x ^{2} + x - 6}{x ^{2} - 9} d x

\int \frac{1}{( 95 - x ) ^{2}} d x

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

\frac{d}{d x} ((4 x - 1)^{3})

What is the integral of x^3sqrt(x^2+41) ?
The integral of x^3sqrt(x^2+41) is 1/5 (x^2+41)^{5/2}-41/3 (x^2+41)^{3/2}+C