What is the integral of (5x+1)/(sqrt(4x^2+4x+2)) ?

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

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integral of (5x+1)/(sqrt(4x^2+4x+2))

\int \frac{5 x + 1}{4 x ^{2} + 4 x + 2} d x

\frac{5}{4} 4 x^{2} + 4 x + 2 - \frac{3}{4} ln 2 x + 1 + 4 x^{2} + 4 x + 2 + C

Solution steps

\int \frac{5 x + 1}{4 x ^{2} + 4 x + 2} d x

Complete the square

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

= \int \frac{5 x + 1}{4 ( x + \frac{1}{2} ) ^{2} + 1} d x

Apply u-substitution

: \int \frac{5 ( u - \frac{1}{2} ) + 1}{4 u ^{2} + 1} d u

= \int \frac{5 ( u - \frac{1}{2} ) + 1}{4 u ^{2} + 1} d u

Expand

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

= \int \frac{5 u}{4 u ^{2} + 1} - \frac{3}{2 4 u ^{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{5 u}{4 u ^{2} + 1} d u - \int \frac{3}{2 4 u ^{2} + 1} d u

\int \frac{5 u}{4 u ^{2} + 1} d u = \frac{5}{4} 4 u^{2} + 1

\int \frac{3}{2 4 u ^{2} + 1} d u = \frac{3}{4} ln 2 u + 1 + 4 u^{2}

= \frac{5}{4} 4 u^{2} + 1 - \frac{3}{4} ln 2 u + 1 + 4 u^{2}

Substitute back

u = x + \frac{1}{2}

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

Simplify

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

= \frac{5}{4} 4 x^{2} + 4 x + 2 - \frac{3}{4} ln 2 x + 1 + 4 x^{2} + 4 x + 2

Add a constant to the solution

= \frac{5}{4} 4 x^{2} + 4 x + 2 - \frac{3}{4} ln 2 x + 1 + 4 x^{2} + 4 x + 2 + C

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