What is the integral of (x^3)/(sqrt(25+x^2)) ?

The integral of (x^3)/(sqrt(25+x^2)) is 1/3 (25+x^2)^{3/2}-25sqrt(25+x^2)+C

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integral of (x^3)/(sqrt(25+x^2))

\int \frac{x ^{3}}{25 + x ^{2}} d x

\frac{1}{3} (25 + x^{2})^{\frac{3}{2}} - 25 25 + x^{2} + C

Solution steps

\int \frac{x ^{3}}{25 + x ^{2}} d x

Apply u-substitution

: \int u^{2} - 25 d u

= \int u^{2} - 25 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^{2} d u - \int 25 d u

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

\int 25 d u = 25 u

= \frac{u ^{3}}{3} - 25 u

Substitute back

u = 25 + x^{2}

= \frac{( 25 + x ^{2} ) ^{3}}{3} - 25 25 + x^{2}

Simplify

\frac{( 25 + x ^{2} ) ^{3}}{3} - 25 25 + x^{2} : \frac{1}{3} (25 + x^{2})^{\frac{3}{2}} - 25 25 + x^{2}

= \frac{1}{3} (25 + x^{2})^{\frac{3}{2}} - 25 25 + x^{2}

Add a constant to the solution

= \frac{1}{3} (25 + x^{2})^{\frac{3}{2}} - 25 25 + x^{2} + C

f (x) = - 3 x^{2} + 12, at x = 3

8 x^{2} y + y^{^{'}} = 10 x^{2}

4 x^{3} - 33 x^{2} + 84 x - 60

\frac{d}{d x} (2 x 16 - x^{2})

\frac{\partial}{\partial x} (2 x^{2} + y^{3})

What is the integral of (x^3)/(sqrt(25+x^2)) ?
The integral of (x^3)/(sqrt(25+x^2)) is 1/3 (25+x^2)^{3/2}-25sqrt(25+x^2)+C