What is the integral of (x^2+3x)/(sqrt(x+4)) ?

The integral of (x^2+3x)/(sqrt(x+4)) is 2(1/5 (x+4)^{5/2}-5/3 (x+4)^{3/2}+4sqrt(x+4))+C

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integral of (x^2+3x)/(sqrt(x+4))

\int \frac{x ^{2} + 3 x}{x + 4} d x

2 (\frac{1}{5} (x + 4)^{\frac{5}{2}} - \frac{5}{3} (x + 4)^{\frac{3}{2}} + 4 x + 4) + C

Solution steps

\int \frac{x ^{2} + 3 x}{x + 4} d x

Apply u-substitution

: \int 2 (u^{4} - 5 u^{2} + 4) d u

= \int 2 (u^{4} - 5 u^{2} + 4) d u

Take the constant out:

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

= 2 \cdot \int u^{4} - 5 u^{2} + 4 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^{4} d u - \int 5 u^{2} d u + \int 4 d u)

\int u^{4} d u = \frac{u ^{5}}{5}

\int 5 u^{2} d u = \frac{5 u ^{3}}{3}

\int 4 d u = 4 u

= 2 (\frac{u ^{5}}{5} - \frac{5 u ^{3}}{3} + 4 u)

Substitute back

u = x + 4

= 2 (\frac{( x + 4 ) ^{5}}{5} - \frac{5 ( x + 4 ) ^{3}}{3} + 4 x + 4)

Simplify

2 (\frac{( x + 4 ) ^{5}}{5} - \frac{5 ( x + 4 ) ^{3}}{3} + 4 x + 4) : 2 (\frac{1}{5} (x + 4)^{\frac{5}{2}} - \frac{5}{3} (x + 4)^{\frac{3}{2}} + 4 x + 4)

= 2 (\frac{1}{5} (x + 4)^{\frac{5}{2}} - \frac{5}{3} (x + 4)^{\frac{3}{2}} + 4 x + 4)

Add a constant to the solution

= 2 (\frac{1}{5} (x + 4)^{\frac{5}{2}} - \frac{5}{3} (x + 4)^{\frac{3}{2}} + 4 x + 4) + C

\int (\frac{e ^{4 x} - 5}{e ^{2 x}}) d x

y^{^{'}} - \frac{5}{x} y = \frac{y ^{5}}{x ^{3}}, y (1) = 1

f (x) = (x^{2} - 5)^{2}

\int_{- 2}^{- 1} (15 x^{- 4}) d x

\int 3 x^{2} x^{3} + 2 d x

What is the integral of (x^2+3x)/(sqrt(x+4)) ?
The integral of (x^2+3x)/(sqrt(x+4)) is 2(1/5 (x+4)^{5/2}-5/3 (x+4)^{3/2}+4sqrt(x+4))+C