What is the integral of x/(sqrt(x+6)) ?

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

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integral of x/(sqrt(x+6))

\int \frac{x}{x + 6} d x

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

Solution steps

\int \frac{x}{x + 6} d x

Apply u-substitution

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

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

Take the constant out:

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

= 2 \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

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

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

\int 6 d u = 6 u

= 2 (\frac{u ^{3}}{3} - 6 u)

Substitute back

u = x + 6

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (e^{x} + 2^{(- x)} + 2 cos (x) - 6)

\frac{d y}{d x} + 8 y = e^{x} y^{- 6}

\frac{2 x d y}{d x} + y = 6 x

\frac{d}{d x} (4 x^{3} - 16 x + 5)

\int \frac{1}{1 + x ^{2} + y ^{2}} d y

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