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

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

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

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

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

Solution steps

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

Complete the square

x^{2} + 2 x + 2 : (x + 1)^{2} + 1

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

Apply u-substitution

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

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

Expand

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

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

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

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

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

Substitute back

u = x + 1

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

Simplify

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

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

Add a constant to the solution

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

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