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

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

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integral of (x+1)/(2x+1)

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

\frac{1}{2} x + \frac{1}{4} + \frac{1}{4} ln ∣ 2 x + 1 ∣ + C

Solution steps

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

Expand

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

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

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

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

= \frac{1}{4} (2 x + 1 - ln ∣ 2 x + 1 ∣) + \frac{1}{2} ln ∣ 2 x + 1 ∣

Simplify

\frac{1}{4} (2 x + 1 - ln ∣ 2 x + 1 ∣) + \frac{1}{2} ln ∣ 2 x + 1 ∣ : \frac{1}{2} x + \frac{1}{4} + \frac{1}{4} ln ∣ 2 x + 1 ∣

= \frac{1}{2} x + \frac{1}{4} + \frac{1}{4} ln ∣ 2 x + 1 ∣

Add a constant to the solution

= \frac{1}{2} x + \frac{1}{4} + \frac{1}{4} ln ∣ 2 x + 1 ∣ + C

\int 5 sin^{2} (x) d x

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\int (x - 4)^{2} x^{2} d x

\frac{\partial}{\partial x} (- 3 sin (2 x))

\frac{d}{d x} (2 a sin (x))

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