What is the integral of ((12x+1))/(sqrt(4x-3)) ?

The integral of ((12x+1))/(sqrt(4x-3)) is 1/2 ((4x-3)^{3/2}+10sqrt(4x-3))+C

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integral of ((12x+1))/(sqrt(4x-3))

\int \frac{( 12 x + 1 )}{4 x - 3} d x

\frac{1}{2} ((4 x - 3)^{\frac{3}{2}} + 10 4 x - 3) + C

Solution steps

\int \frac{12 x + 1}{4 x - 3} d x

Apply u-substitution

: \int \frac{3 u ^{2} + 10}{2} d u

= \int \frac{3 u ^{2} + 10}{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 3 u^{2} + 10 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 3 u^{2} d u + \int 10 d u)

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

\int 10 d u = 10 u

= \frac{1}{2} (u^{3} + 10 u)

Substitute back

u = 4 x - 3

= \frac{1}{2} ((4 x - 3)^{3} + 10 4 x - 3)

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

= \frac{1}{2} ((4 x - 3)^{\frac{3}{2}} + 10 4 x - 3)

Add a constant to the solution

= \frac{1}{2} ((4 x - 3)^{\frac{3}{2}} + 10 4 x - 3) + C

\int \frac{1}{( x ^{2} + 2 x + 10 ) ^{2}} d x

\int \frac{4 x ^{3}}{( x ^{2} + 1 ) ^{3}} d x

\int \frac{4}{x ^{- 3}} d x

\int \frac{3 x ^{2} - 7 x + 5}{( x - 1 ) ( x - 2 ) ^{2}} d x

What is the integral of ((12x+1))/(sqrt(4x-3)) ?
The integral of ((12x+1))/(sqrt(4x-3)) is 1/2 ((4x-3)^{3/2}+10sqrt(4x-3))+C