What is the integral of (5x^3)/(sqrt(6+x^2)) ?

The integral of (5x^3)/(sqrt(6+x^2)) is 5(1/3 (6+x^2)^{3/2}-6sqrt(6+x^2))+C

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integral of (5x^3)/(sqrt(6+x^2))

\int \frac{5 x ^{3}}{6 + x ^{2}} d x

5 (\frac{1}{3} (6 + x^{2})^{\frac{3}{2}} - 6 6 + x^{2}) + C

Solution steps

\int \frac{5 x ^{3}}{6 + x ^{2}} d x

Take the constant out:

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

= 5 \cdot \int \frac{x ^{3}}{6 + x ^{2}} d x

Apply u-substitution

: \int u^{2} - 6 d u

= 5 \cdot \int u^{2} - 6 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

= 5 (\int u^{2} d u - \int 6 d u)

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

\int 6 d u = 6 u

= 5 (\frac{u ^{3}}{3} - 6 u)

Substitute back

u = 6 + x^{2}

= 5 (\frac{( 6 + x ^{2} ) ^{3}}{3} - 6 6 + x^{2})

Simplify

5 (\frac{( 6 + x ^{2} ) ^{3}}{3} - 6 6 + x^{2}) : 5 (\frac{1}{3} (6 + x^{2})^{\frac{3}{2}} - 6 6 + x^{2})

= 5 (\frac{1}{3} (6 + x^{2})^{\frac{3}{2}} - 6 6 + x^{2})

Add a constant to the solution

= 5 (\frac{1}{3} (6 + x^{2})^{\frac{3}{2}} - 6 6 + x^{2}) + C

\frac{d}{d x} (2 sin (2 x))

\frac{d ^{4}}{d x ^{4}} (\frac{1}{x ^{2}})

\int x^{2} cos (mπ x) d x

sin (5 x) + cos (4 x)

f (x) = \frac{1}{x + 2}, at x = 0

What is the integral of (5x^3)/(sqrt(6+x^2)) ?
The integral of (5x^3)/(sqrt(6+x^2)) is 5(1/3 (6+x^2)^{3/2}-6sqrt(6+x^2))+C