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

The integral of (sqrt(2x+1))/(x+1) is 2(-arctan(sqrt(2x+1))+sqrt(2x+1))+C

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

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

2 (- arctan (2 x + 1) + 2 x + 1) + C

Solution steps

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

Apply u-substitution

: \int \frac{u}{u + 1} d u

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

Apply u-substitution

: \int \frac{2 v ^{2}}{v ^{2} + 1} d v

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{v ^{2}}{v ^{2} + 1} d v

\frac{v ^{2}}{v ^{2} + 1} = - \frac{1}{v ^{2} + 1} + 1

= 2 \cdot \int - \frac{1}{v ^{2} + 1} + 1 d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

\int 1 d v = v

= 2 (- arctan (v) + v)

Substitute back

= 2 (- arctan (2 x + 1) + 2 x + 1)

Add a constant to the solution

= 2 (- arctan (2 x + 1) + 2 x + 1) + C

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

y^{^{'}} = \frac{2 x}{1 + 2 y}, y (0) = 2

\int sec (\frac{x}{5}) d x

\frac{1}{s ( s ^{2} + 5 s + 6 )}

x \to 0 lim (x sin (\frac{1}{x ^{2}}))

What is the integral of (sqrt(2x+1))/(x+1) ?
The integral of (sqrt(2x+1))/(x+1) is 2(-arctan(sqrt(2x+1))+sqrt(2x+1))+C