What is the integral of (2x)/(sqrt(4x-5)) ?

The integral of (2x)/(sqrt(4x-5)) is 1/6 (2x+5)sqrt(4x-5)+C

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integral of (2x)/(sqrt(4x-5))

\int \frac{2 x}{4 x - 5} d x

\frac{1}{6} (2 x + 5) 4 x - 5 + C

Solution steps

\int \frac{2 x}{4 x - 5} d x

Take the constant out:

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

= 2 \cdot \int \frac{x}{4 x - 5} d x

Apply u-substitution

: \int \frac{u ^{2} + 5}{8} d u

= 2 \cdot \int \frac{u ^{2} + 5}{8} d u

Take the constant out:

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

= 2 \cdot \frac{1}{8} \cdot \int u^{2} + 5 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 \cdot \frac{1}{8} (\int u^{2} d u + \int 5 d u)

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

\int 5 d u = 5 u

= 2 \cdot \frac{1}{8} (\frac{u ^{3}}{3} + 5 u)

Substitute back

u = 4 x - 5

= 2 \cdot \frac{1}{8} (\frac{( 4 x - 5 ) ^{3}}{3} + 5 4 x - 5)

Simplify

2 \cdot \frac{1}{8} (\frac{( 4 x - 5 ) ^{3}}{3} + 5 4 x - 5) : \frac{1}{6} (2 x + 5) 4 x - 5

= \frac{1}{6} (2 x + 5) 4 x - 5

Add a constant to the solution

= \frac{1}{6} (2 x + 5) 4 x - 5 + C

\frac{d r}{d t} = \frac{r ^{2}}{t}

(- 3 x + 2)^{2}

\frac{d}{d x} (5^{3 x} + 7)

(\frac{x + 7}{x - 7})^{5}

y^{^{'}} + (\frac{3}{300 - x}) y = 4

What is the integral of (2x)/(sqrt(4x-5)) ?
The integral of (2x)/(sqrt(4x-5)) is 1/6 (2x+5)sqrt(4x-5)+C