What is the integral from 0 to 3 of (x+1)^{1/2} ?

The integral from 0 to 3 of (x+1)^{1/2} is 14/3

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integral from 0 to 3 of (x+1)^{1/2}

\int_{0}^{3} (x + 1)^{\frac{1}{2}} d x

\frac{14}{3}

Decimal Notation

4.66666 \dots

Solution steps

\int_{0}^{3} (x + 1)^{\frac{1}{2}} d x

Multiply by the conjugate of

x + 1 : \frac{x ^{2} - 1}{x - 1}

= \int_{0}^{3} (\frac{x ^{2} - 1}{x - 1})^{\frac{1}{2}} d x

Find undefined (singularity) points:

x = 1

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{0}^{1} (\frac{x ^{2} - 1}{x - 1})^{\frac{1}{2}} d x + \int_{1}^{3} (\frac{x ^{2} - 1}{x - 1})^{\frac{1}{2}} d x

\int_{0}^{1} (\frac{x ^{2} - 1}{x - 1})^{\frac{1}{2}} d x = \frac{4 \cdot 2 ^{\frac{1}{2}} - 2}{3}

\int_{1}^{3} (\frac{x ^{2} - 1}{x - 1})^{\frac{1}{2}} d x = \frac{16 - 4 \cdot 2 ^{\frac{1}{2}}}{3}

= \frac{4 \cdot 2 ^{\frac{1}{2}} - 2}{3} + \frac{16 - 4 \cdot 2 ^{\frac{1}{2}}}{3}

Simplify

\frac{4 \cdot 2 ^{\frac{1}{2}} - 2}{3} + \frac{16 - 4 \cdot 2 ^{\frac{1}{2}}}{3} : \frac{14}{3}

= \frac{14}{3}

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\int \frac{e ^{x}}{49 - e ^{2 x}} d x

\frac{d}{d x} (x + \frac{1}{2})

What is the integral from 0 to 3 of (x+1)^{1/2} ?
The integral from 0 to 3 of (x+1)^{1/2} is 14/3