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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

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

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

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

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

\int 1 d u = u

= 2 (- arctan (u) + u)

Substitute back

u = x

= 2 (- arctan (x) + x)

Add a constant to the solution

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

f (x) = 3 x + 1, at x = 8

\int 3 1 - x^{2} d x

p (t) = 25 e^{0.03 t}

\int \frac{( x ^{3} - 1 )}{( 4 x ^{3} - x )} d x

\int \frac{- 4 x - 6}{x ^{2} + x + 1} d x

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