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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u}{2 ( - u ^{2} + 4 u ) ^{\frac{3}{2}}} d u

= \int \frac{u}{2 ( - u ^{2} + 4 u ) ^{\frac{3}{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 \frac{u}{( - u ^{2} + 4 u ) ^{\frac{3}{2}}} d u

Complete the square

- u^{2} + 4 u : - (u - 2)^{2} + 4

= \frac{1}{2} \cdot \int \frac{u}{( - ( u - 2 ) ^{2} + 4 ) ^{\frac{3}{2}}} d u

Apply u-substitution

: \int \frac{v + 2}{( - v ^{2} + 4 ) ^{\frac{3}{2}}} d v

= \frac{1}{2} \cdot \int \frac{v + 2}{( - v ^{2} + 4 ) ^{\frac{3}{2}}} d v

Expand

\frac{v + 2}{( - v ^{2} + 4 ) ^{\frac{3}{2}}} : \frac{v}{( - v ^{2} + 4 ) ^{\frac{3}{2}}} + \frac{2}{( - v ^{2} + 4 ) ^{\frac{3}{2}}}

= \frac{1}{2} \cdot \int \frac{v}{( - v ^{2} + 4 ) ^{\frac{3}{2}}} + \frac{2}{( - v ^{2} + 4 ) ^{\frac{3}{2}}} d v

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

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

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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