What is the integral of ((3x-1))/(sqrt(2x+1)) ?

The integral of ((3x-1))/(sqrt(2x+1)) is 1/2 (2x+1)^{3/2}-5/2 sqrt(2x+1)+C

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integral of ((3x-1))/(sqrt(2x+1))

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

\frac{1}{2} (2 x + 1)^{\frac{3}{2}} - \frac{5}{2} 2 x + 1 + C

Solution steps

\int \frac{3 x - 1}{2 x + 1} d x

Apply u-substitution

: \int \frac{3 ( u ^{2} - 1 )}{2} - 1 d u

= \int \frac{3 ( u ^{2} - 1 )}{2} - 1 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

= \int \frac{3 ( u ^{2} - 1 )}{2} d u - \int 1 d u

\int \frac{3 ( u ^{2} - 1 )}{2} d u = \frac{u ^{3}}{2} - \frac{3}{2} u

\int 1 d u = u

= \frac{u ^{3}}{2} - \frac{3}{2} u - u

Substitute back

u = 2 x + 1

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

Simplify

\frac{( 2 x + 1 ) ^{3}}{2} - \frac{3}{2} 2 x + 1 - 2 x + 1 : \frac{1}{2} (2 x + 1)^{\frac{3}{2}} - \frac{5}{2} 2 x + 1

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

Add a constant to the solution

= \frac{1}{2} (2 x + 1)^{\frac{3}{2}} - \frac{5}{2} 2 x + 1 + C

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What is the integral of ((3x-1))/(sqrt(2x+1)) ?
The integral of ((3x-1))/(sqrt(2x+1)) is 1/2 (2x+1)^{3/2}-5/2 sqrt(2x+1)+C