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

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

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

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

2 x + 1 + C

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

: 2 u^{\frac{1}{2}}

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

Substitute back

u = x + 1

= 2 (x + 1)^{\frac{1}{2}}

Apply radical rule:

= 2 x + 1

Add a constant to the solution

= 2 x + 1 + C

\int \frac{\frac{5 ( x ^{3} + 2 )}{x ^{4}}}{\frac{- 30}{x}} d x

(\frac{1}{3}) x^{3} - 5 x

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

\int \frac{5 x ^{4} - 8 x ^{2} + 3 x - 2}{x ^{2}} d x

\int \frac{x - 1}{x ^{3} + 1} d x

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