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

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

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integral of x/(sqrt(x+7))

\int \frac{x}{x + 7} d x

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

Solution steps

\int \frac{x}{x + 7} d x

Apply u-substitution

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

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

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

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

\int 7 d u = 7 u

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

Substitute back

u = x + 7

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

Simplify

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

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

Add a constant to the solution

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

y^{^{''}} + 4 y = 4 sec (2 t)

x y^{^{'}} - y = 2020 x^{2}

(- 7, - 4), (1, - 6)

2 x^{2}, 2 x + 6, - 2, 2

y^{^{''}} - y^{^{'}} - 30 y = 0, y (0) = 3, y^{^{'}} (0) = - 3

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