What is the integral of (12)/(3x+xsqrt(x)) ?

The integral of (12)/(3x+xsqrt(x)) is 24(1/3 ln|sqrt(x)|-1/3 ln|sqrt(x)+3|)+C

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integral of (12)/(3x+xsqrt(x))

\int \frac{12}{3 x + x x} d x

24 (\frac{1}{3} ln x - \frac{1}{3} ln x + 3) + C

Solution steps

\int \frac{12}{3 x + x x} d x

Take the constant out:

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

= 12 \cdot \int \frac{1}{3 x + x x} d x

Apply u-substitution

: \int \frac{2}{u ( u + 3 )} d u

= 12 \cdot \int \frac{2}{u ( u + 3 )} d u

Take the constant out:

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

= 12 \cdot 2 \cdot \int \frac{1}{u ( u + 3 )} d u

Take the partial fraction of

\frac{1}{u ( u + 3 )} : \frac{1}{3 u} - \frac{1}{3 ( u + 3 )}

= 12 \cdot 2 \cdot \int \frac{1}{3 u} - \frac{1}{3 ( u + 3 )} 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

= 12 \cdot 2 (\int \frac{1}{3 u} d u - \int \frac{1}{3 ( u + 3 )} d u)

\int \frac{1}{3 u} d u = \frac{1}{3} ln ∣ u ∣

\int \frac{1}{3 ( u + 3 )} d u = \frac{1}{3} ln ∣ u + 3 ∣

= 12 \cdot 2 (\frac{1}{3} ln ∣ u ∣ - \frac{1}{3} ln ∣ u + 3 ∣)

Substitute back

u = x

= 12 \cdot 2 (\frac{1}{3} ln x - \frac{1}{3} ln x + 3)

Simplify

= 24 (\frac{1}{3} ln x - \frac{1}{3} ln x + 3)

Add a constant to the solution

= 24 (\frac{1}{3} ln x - \frac{1}{3} ln x + 3) + C

(3 + 3 x^{2}) \frac{d y}{d x} + 2 x y = 2 x y^{4}

f (x) = ln (2)

\int 3 e^{4} d x

2000 (1 + \frac{r}{12})^{60}

f (x) = \frac{1}{x}, (a, \frac{1}{a})

What is the integral of (12)/(3x+xsqrt(x)) ?
The integral of (12)/(3x+xsqrt(x)) is 24(1/3 ln|sqrt(x)|-1/3 ln|sqrt(x)+3|)+C