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

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

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

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

2 x^{2} + 4 x + 5 - 5 ln x^{2} + 4 x + 5 + x + 2 + C

Solution steps

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

Complete the square

x^{2} + 4 x + 5 : (x + 2)^{2} + 1

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

Apply u-substitution

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

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

Expand

\frac{2 u - 5}{u ^{2} + 1} : \frac{2 u}{u ^{2} + 1} - \frac{5}{u ^{2} + 1}

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

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

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

= 2 u^{2} + 1 - 5 ln u^{2} + 1 + u

Substitute back

u = x + 2

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

Simplify

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

= 2 x^{2} + 4 x + 5 - 5 ln x^{2} + 4 x + 5 + x + 2

Add a constant to the solution

= 2 x^{2} + 4 x + 5 - 5 ln x^{2} + 4 x + 5 + x + 2 + C

\frac{\partial}{\partial z} (2 x y - z^{2})

\int x \cdot e^{\frac{x ^{2}}{2}} d x

- 9 e^{x} sin (x) + 9 e^{x} cos (x)

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

\frac{d}{d θ} (- 3 sin (θ))

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