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

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

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

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

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

Solution steps

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

Expand

\frac{3 x + 4}{2 x + x ^{2}} : \frac{3 x}{2 x + x ^{2}} + \frac{4}{2 x + x ^{2}}

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\int 3 x - 1 d x

\frac{y ^{3}}{3} + \frac{1}{4 y}

\int_{4}^{5} \frac{x - 4}{x} d x

\frac{\partial}{\partial x} (x^{2} + 3 x y + y^{2} + 1)

\int \frac{1}{3 - 2 y} d y

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