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

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

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

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

2 (- arctan (n) + n) + C

Solution steps

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

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 = n

= 2 (- arctan (n) + n)

Add a constant to the solution

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

\int (3 x + \frac{10}{x ^{6}}) d x

f (x) = 3 x^{\frac{1}{2}} + x^{\frac{3}{2}}, at x = 9

\int 6 sin (2 t) d t

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

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

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