What is the integral from 0 to 5 of 1/(sqrt(9-2x)) ?

The integral from 0 to 5 of 1/(sqrt(9-2x)) is 3-i

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integral from 0 to 5 of 1/(sqrt(9-2x))

\int_{0}^{5} \frac{1}{9 - 2 x} d x

3 - i

Solution steps

\int_{0}^{5} \frac{1}{9 - 2 x} d x

Adjust integral based on undefined pointsThere is an undefined point within the boundaries at

: \frac{9}{2}

= \int_{0}^{\frac{9}{2}} \frac{1}{9 - 2 x} d x + \int_{\frac{9}{2}}^{5} \frac{1}{9 - 2 x} d x

\int_{0}^{\frac{9}{2}} \frac{1}{9 - 2 x} d x = 3

= 3 + \int_{\frac{9}{2}}^{5} \frac{1}{9 - 2 x} d x

\int_{\frac{9}{2}}^{5} \frac{1}{9 - 2 x} d x = - i

= 3 - i

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What is the integral from 0 to 5 of 1/(sqrt(9-2x)) ?
The integral from 0 to 5 of 1/(sqrt(9-2x)) is 3-i