What is the integral of x/(sqrt(x+20)) ?

The integral of x/(sqrt(x+20)) is 2(1/3 (x+20)^{3/2}-20sqrt(x+20))+C

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integral of x/(sqrt(x+20))

\int \frac{x}{x + 20} d x

2 (\frac{1}{3} (x + 20)^{\frac{3}{2}} - 20 x + 20) + C

Solution steps

\int \frac{x}{x + 20} d x

Apply u-substitution

: \int 2 (u^{2} - 20) d u

= \int 2 (u^{2} - 20) 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} - 20 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 20 d u)

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

\int 20 d u = 20 u

= 2 (\frac{u ^{3}}{3} - 20 u)

Substitute back

u = x + 20

= 2 (\frac{( x + 20 ) ^{3}}{3} - 20 x + 20)

Simplify

2 (\frac{( x + 20 ) ^{3}}{3} - 20 x + 20) : 2 (\frac{1}{3} (x + 20)^{\frac{3}{2}} - 20 x + 20)

= 2 (\frac{1}{3} (x + 20)^{\frac{3}{2}} - 20 x + 20)

Add a constant to the solution

= 2 (\frac{1}{3} (x + 20)^{\frac{3}{2}} - 20 x + 20) + C

(x + x) (\frac{d y}{d x}) = y + y

f (x) = cos (x), at (0)

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

4 x^{2} - 2 x + 3

\int_{0}^{π} sin (x) + cos (2 x) d x

What is the integral of x/(sqrt(x+20)) ?
The integral of x/(sqrt(x+20)) is 2(1/3 (x+20)^{3/2}-20sqrt(x+20))+C