What is the integral of (28)/(4x+xsqrt(x)) ?

The integral of (28)/(4x+xsqrt(x)) is 56(1/4 ln|sqrt(x)|-1/4 ln|sqrt(x)+4|)+C

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integral of (28)/(4x+xsqrt(x))

\int \frac{28}{4 x + x x} d x

56 (\frac{1}{4} ln x - \frac{1}{4} ln x + 4) + C

Solution steps

\int \frac{28}{4 x + x x} d x

Take the constant out:

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

= 28 \cdot \int \frac{1}{4 x + x x} d x

Apply u-substitution

: \int \frac{2}{u ( u + 4 )} d u

= 28 \cdot \int \frac{2}{u ( u + 4 )} d u

Take the constant out:

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

= 28 \cdot 2 \cdot \int \frac{1}{u ( u + 4 )} d u

Take the partial fraction of

\frac{1}{u ( u + 4 )} : \frac{1}{4 u} - \frac{1}{4 ( u + 4 )}

= 28 \cdot 2 \cdot \int \frac{1}{4 u} - \frac{1}{4 ( u + 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

= 28 \cdot 2 (\int \frac{1}{4 u} d u - \int \frac{1}{4 ( u + 4 )} d u)

\int \frac{1}{4 u} d u = \frac{1}{4} ln ∣ u ∣

\int \frac{1}{4 ( u + 4 )} d u = \frac{1}{4} ln ∣ u + 4 ∣

= 28 \cdot 2 (\frac{1}{4} ln ∣ u ∣ - \frac{1}{4} ln ∣ u + 4 ∣)

Substitute back

u = x

= 28 \cdot 2 (\frac{1}{4} ln x - \frac{1}{4} ln x + 4)

Simplify

= 56 (\frac{1}{4} ln x - \frac{1}{4} ln x + 4)

Add a constant to the solution

= 56 (\frac{1}{4} ln x - \frac{1}{4} ln x + 4) + C

\int_{0}^{2} y^{2} 1 + y^{3} d y

x \to - \frac{1}{2} lim (\frac{6 x - 1}{x})

\int x^{2} 2 x - x^{2} d x

f (x) = x^{3} - 5 x^{2}, at x = 2

\frac{d y}{d x} = \frac{x ^{2} + x y + y ^{2}}{x ^{2}}

What is the integral of (28)/(4x+xsqrt(x)) ?
The integral of (28)/(4x+xsqrt(x)) is 56(1/4 ln|sqrt(x)|-1/4 ln|sqrt(x)+4|)+C