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

The integral from 2 to 3 of 1/(|x-3|^{1/2)} is 2

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integral from 2 to 3 of 1/(|x-3|^{1/2)}

\int_{2}^{3} \frac{1}{∣ x - 3 ∣ ^{\frac{1}{2}}} d x

2

Solution steps

\int_{2}^{3} \frac{1}{∣ x - 3 ∣ ^{\frac{1}{2}}} d x

Eliminate Absolutes

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

Apply Integration By Parts

: [\frac{x}{( - x + 3 ) ^{\frac{1}{2}}} - \int \frac{x}{2 ( - x + 3 ) ^{\frac{3}{2}}} d x]_{2}^{3}

= [\frac{x}{( - x + 3 ) ^{\frac{1}{2}}} - \int \frac{x}{2 ( - x + 3 ) ^{\frac{3}{2}}} d x]_{2}^{3}

\int \frac{x}{2 ( - x + 3 ) ^{\frac{3}{2}}} d x = - \frac{x - 6}{( - x + 3 ) ^{\frac{1}{2}}}

= [\frac{x}{( - x + 3 ) ^{\frac{1}{2}}} - (- \frac{x - 6}{( - x + 3 ) ^{\frac{1}{2}}})]_{2}^{3}

Simplify

[\frac{x}{( - x + 3 ) ^{\frac{1}{2}}} - (- \frac{x - 6}{( - x + 3 ) ^{\frac{1}{2}}})]_{2}^{3} : [- 2 (- x + 3)^{\frac{1}{2}}]_{2}^{3}

= [- 2 (- x + 3)^{\frac{1}{2}}]_{2}^{3}

Compute the boundaries:

2

= 2

\int (\frac{x}{6} + \frac{6}{x}) d x

\int x^{2} x^{2} - 4 d x

\frac{d}{d x} (\frac{8}{5 x ^{4}})

y^{^{''}} + 25 y = csc^{2} (5 t)

(- 1, - 6), (- 15, 2)

What is the integral from 2 to 3 of 1/(|x-3|^{1/2)} ?
The integral from 2 to 3 of 1/(|x-3|^{1/2)} is 2