What is the integral of (8x+1)/(sqrt(4x-3)) ?

The integral of (8x+1)/(sqrt(4x-3)) is 1/3 (4x-3)^{3/2}+7/2 sqrt(4x-3)+C

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integral of (8x+1)/(sqrt(4x-3))

\int \frac{8 x + 1}{4 x - 3} d x

\frac{1}{3} (4 x - 3)^{\frac{3}{2}} + \frac{7}{2} 4 x - 3 + C

Solution steps

\int \frac{8 x + 1}{4 x - 3} d x

Apply u-substitution

= \int \frac{2 u ^{2} + 7}{2} d u

Take the constant out:

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

= \frac{1}{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

= \frac{1}{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

= \frac{1}{2} (\frac{2 u ^{3}}{3} + 7 u)

Substitute back

u = 4 x - 3

= \frac{1}{2} (\frac{2 ( 4 x - 3 ) ^{3}}{3} + 7 4 x - 3)

Simplify

\frac{1}{2} (\frac{2 ( 4 x - 3 ) ^{3}}{3} + 7 4 x - 3) : \frac{1}{3} (4 x - 3)^{\frac{3}{2}} + \frac{7}{2} 4 x - 3

= \frac{1}{3} (4 x - 3)^{\frac{3}{2}} + \frac{7}{2} 4 x - 3

Add a constant to the solution

= \frac{1}{3} (4 x - 3)^{\frac{3}{2}} + \frac{7}{2} 4 x - 3 + C

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What is the integral of (8x+1)/(sqrt(4x-3)) ?
The integral of (8x+1)/(sqrt(4x-3)) is 1/3 (4x-3)^{3/2}+7/2 sqrt(4x-3)+C