What is the integral of (2x+3)/(sqrt(4x^2-4x+5)) ?

The integral of (2x+3)/(sqrt(4x^2-4x+5)) is 1/2 (sqrt(4x^2-4x+5)+4ln((|2x+sqrt(4x^2-4x+5)-1|)/2))+C

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integral of (2x+3)/(sqrt(4x^2-4x+5))

\int \frac{2 x + 3}{4 x ^{2} - 4 x + 5} d x

\frac{1}{2} (4 x^{2} - 4 x + 5 + 4 ln (\frac{2 x + 4 x ^{2} - 4 x + 5 - 1}{2})) + C

Solution steps

\int \frac{2 x + 3}{4 x ^{2} - 4 x + 5} d x

Apply u-substitution

: \int \frac{u}{2 u ^{2} - 8 u + 20} d u

= \int \frac{u}{2 u ^{2} - 8 u + 20} 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} - 8 u + 20} d u

Complete the square

u^{2} - 8 u + 20 : (u - 4)^{2} + 4

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

Apply u-substitution

: \int \frac{v + 4}{v ^{2} + 4} d v

= \frac{1}{2} \cdot \int \frac{v + 4}{v ^{2} + 4} d v

Expand

\frac{v + 4}{v ^{2} + 4} : \frac{v}{v ^{2} + 4} + \frac{4}{v ^{2} + 4}

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

\int \frac{v}{v ^{2} + 4} d v = v^{2} + 4

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

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

Substitute back

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

Simplify

\frac{1}{2} ((2 x + 3 - 4)^{2} + 4 + 4 ln (\frac{1}{2} 2 x + 3 - 4 + 4 + (2 x + 3 - 4)^{2})) : \frac{1}{2} (4 x^{2} - 4 x + 5 + 4 ln (\frac{2 x + 4 x ^{2} - 4 x + 5 - 1}{2}))

= \frac{1}{2} (4 x^{2} - 4 x + 5 + 4 ln (\frac{2 x + 4 x ^{2} - 4 x + 5 - 1}{2}))

Add a constant to the solution

= \frac{1}{2} (4 x^{2} - 4 x + 5 + 4 ln (\frac{2 x + 4 x ^{2} - 4 x + 5 - 1}{2})) + C

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What is the integral of (2x+3)/(sqrt(4x^2-4x+5)) ?
The integral of (2x+3)/(sqrt(4x^2-4x+5)) is 1/2 (sqrt(4x^2-4x+5)+4ln((|2x+sqrt(4x^2-4x+5)-1|)/2))+C