What is the integral of ((4x-1))/((2x+1)^{3/2)} ?

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

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integral of ((4x-1))/((2x+1)^{3/2)}

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

Simplify

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

Add a constant to the solution

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

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