What is the integral of (x^3)/(sqrt(x^2+7)) ?

The integral of (x^3)/(sqrt(x^2+7)) is 1/3 (x^2+7)^{3/2}-7sqrt(x^2+7)+C

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integral of (x^3)/(sqrt(x^2+7))

\int \frac{x ^{3}}{x ^{2} + 7} d x

\frac{1}{3} (x^{2} + 7)^{\frac{3}{2}} - 7 x^{2} + 7 + C

Solution steps

\int \frac{x ^{3}}{x ^{2} + 7} d x

Apply u-substitution

: \int u^{2} - 7 d u

= \int u^{2} - 7 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 7 d u

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

\int 7 d u = 7 u

= \frac{u ^{3}}{3} - 7 u

Substitute back

u = x^{2} + 7

= \frac{( x ^{2} + 7 ) ^{3}}{3} - 7 x^{2} + 7

Simplify

\frac{( x ^{2} + 7 ) ^{3}}{3} - 7 x^{2} + 7 : \frac{1}{3} (x^{2} + 7)^{\frac{3}{2}} - 7 x^{2} + 7

= \frac{1}{3} (x^{2} + 7)^{\frac{3}{2}} - 7 x^{2} + 7

Add a constant to the solution

= \frac{1}{3} (x^{2} + 7)^{\frac{3}{2}} - 7 x^{2} + 7 + C

f (0.6) = 3 arcsin (x^{4})

\int y e^{x^{2}} d x

\frac{\partial}{\partial x} (\frac{3 x ^{2}}{y - 5})

\frac{\partial}{\partial z} (\frac{x + y}{z})

ln (1 - e^{- x})

What is the integral of (x^3)/(sqrt(x^2+7)) ?
The integral of (x^3)/(sqrt(x^2+7)) is 1/3 (x^2+7)^{3/2}-7sqrt(x^2+7)+C