What is the integral of ((2x+1))/(sqrt(x+4)) ?

The integral of ((2x+1))/(sqrt(x+4)) is 2(2/3 (x+4)^{3/2}-7sqrt(x+4))+C

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integral of ((2x+1))/(sqrt(x+4))

\int \frac{( 2 x + 1 )}{x + 4} d x

2 (\frac{2}{3} (x + 4)^{\frac{3}{2}} - 7 x + 4) + C

Solution steps

\int \frac{2 x + 1}{x + 4} d x

Apply u-substitution

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

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

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

\int 7 d u = 7 u

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

Substitute back

u = x + 4

= 2 (\frac{2 ( x + 4 ) ^{3}}{3} - 7 x + 4)

Simplify

2 (\frac{2 ( x + 4 ) ^{3}}{3} - 7 x + 4) : 2 (\frac{2}{3} (x + 4)^{\frac{3}{2}} - 7 x + 4)

= 2 (\frac{2}{3} (x + 4)^{\frac{3}{2}} - 7 x + 4)

Add a constant to the solution

= 2 (\frac{2}{3} (x + 4)^{\frac{3}{2}} - 7 x + 4) + C

\frac{d}{d x} (e^{x^{3}})

\int_{- \infty}^{0} e^{7 x} d x

\int e^{x} (e - 1) d x

\int \frac{7 x ^{3} + 8 x}{x - 1} d x

(ln (x + x^{2} + 1))^{^{'}}

What is the integral of ((2x+1))/(sqrt(x+4)) ?
The integral of ((2x+1))/(sqrt(x+4)) is 2(2/3 (x+4)^{3/2}-7sqrt(x+4))+C