What is the integral of (14)/(2x+xsqrt(x)) ?

The integral of (14)/(2x+xsqrt(x)) is 28(1/2 ln|sqrt(x)|-1/2 ln|sqrt(x)+2|)+C

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integral of (14)/(2x+xsqrt(x))

\int \frac{14}{2 x + x x} d x

28 (\frac{1}{2} ln x - \frac{1}{2} ln x + 2) + C

Solution steps

\int \frac{14}{2 x + x x} d x

Take the constant out:

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

= 14 \cdot \int \frac{1}{2 x + x x} d x

Apply u-substitution

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

= 14 \cdot \int \frac{2}{u ( u + 2 )} d u

Take the constant out:

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

= 14 \cdot 2 \cdot \int \frac{1}{u ( u + 2 )} d u

Take the partial fraction of

\frac{1}{u ( u + 2 )} : \frac{1}{2 u} - \frac{1}{2 ( u + 2 )}

= 14 \cdot 2 \cdot \int \frac{1}{2 u} - \frac{1}{2 ( u + 2 )} 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

= 14 \cdot 2 (\int \frac{1}{2 u} d u - \int \frac{1}{2 ( u + 2 )} d u)

\int \frac{1}{2 u} d u = \frac{1}{2} ln ∣ u ∣

\int \frac{1}{2 ( u + 2 )} d u = \frac{1}{2} ln ∣ u + 2 ∣

= 14 \cdot 2 (\frac{1}{2} ln ∣ u ∣ - \frac{1}{2} ln ∣ u + 2 ∣)

Substitute back

u = x

= 14 \cdot 2 (\frac{1}{2} ln x - \frac{1}{2} ln x + 2)

Simplify

= 28 (\frac{1}{2} ln x - \frac{1}{2} ln x + 2)

Add a constant to the solution

= 28 (\frac{1}{2} ln x - \frac{1}{2} ln x + 2) + C

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y^{^{''}} + y^{^{'}} + y = 1

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What is the integral of (14)/(2x+xsqrt(x)) ?
The integral of (14)/(2x+xsqrt(x)) is 28(1/2 ln|sqrt(x)|-1/2 ln|sqrt(x)+2|)+C