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

The integral of sqrt(x+1) is 2/3 (x+1)^{3/2}+C

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integral of sqrt(x+1)

\int x + 1 d x

\frac{2}{3} (x + 1)^{\frac{3}{2}} + C

Solution steps

\int x + 1 d x

Apply u-substitution

: \int u d u

= \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = x + 1

= \frac{2}{3} (x + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{3} (x + 1)^{\frac{3}{2}} + C

\int sec^{2} (x) + sec (x) tan (x) d x

\int \frac{π}{2} (x + cos (x)) d x

\frac{d}{d x} (x^{2} - 4 x + 1)

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\int x 2 x^{2} + 7 d x

What is the integral of sqrt(x+1) ?
The integral of sqrt(x+1) is 2/3 (x+1)^{3/2}+C