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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\int 31 sec^{- 2} (x) tan^{3} (x) d x

\int (y + y cos (x y)) d x

\int sin (θ) tan (θ) d θ

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