What is the integral of (11x^3)/(sqrt(7+x^2)) ?

The integral of (11x^3)/(sqrt(7+x^2)) is 11(1/3 (7+x^2)^{3/2}-7sqrt(7+x^2))+C

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integral of (11x^3)/(sqrt(7+x^2))

\int \frac{11 x ^{3}}{7 + x ^{2}} d x

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

Solution steps

\int \frac{11 x ^{3}}{7 + x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 11 \cdot \int \frac{x ^{3}}{7 + x ^{2}} d x

Apply u-substitution

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

= 11 \cdot \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

= 11 (\int u^{2} d u - \int 7 d u)

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

\int 7 d u = 7 u

= 11 (\frac{u ^{3}}{3} - 7 u)

Substitute back

u = 7 + x^{2}

= 11 (\frac{( 7 + x ^{2} ) ^{3}}{3} - 7 7 + x^{2})

Simplify

11 (\frac{( 7 + x ^{2} ) ^{3}}{3} - 7 7 + x^{2}) : 11 (\frac{1}{3} (7 + x^{2})^{\frac{3}{2}} - 7 7 + x^{2})

= 11 (\frac{1}{3} (7 + x^{2})^{\frac{3}{2}} - 7 7 + x^{2})

Add a constant to the solution

= 11 (\frac{1}{3} (7 + x^{2})^{\frac{3}{2}} - 7 7 + x^{2}) + C

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

\frac{d}{d x} ((1 + e^{- x})^{- 1})

y = x^{2 x + 5}

\frac{\partial}{\partial x} (ln (9 - x^{2} - y^{2}))

\frac{\partial}{\partial z} (z (y - yz)^{4})

What is the integral of (11x^3)/(sqrt(7+x^2)) ?
The integral of (11x^3)/(sqrt(7+x^2)) is 11(1/3 (7+x^2)^{3/2}-7sqrt(7+x^2))+C