int_{3}^{5} frac{1}{2x + 3} dx is equal to | vedclass

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(B) To evaluate the integral $I = \int_{3}^{5} \frac{1}{2x + 3} dx$,we use the substitution method.
Let $u = 2x + 3$,then $du = 2 dx$,which implies $

Vedclass · Accepted Answer

$\frac{1}{2} \ln \left( \frac{13}{9} \right)$

Vedclass · Answer

(B) To evaluate the integral $I = \int_{3}^{5} \frac{1}{2x + 3} dx$,we use the substitution method.
Let $u = 2x + 3$,then $du = 2 dx$,which implies $dx = \frac{du}{2}$.
When $x = 3$,$u = 2(3) + 3 = 9$.
When $x = 5$,$u = 2(5) + 3 = 13$.
Substituting these into the integral:
$I = \int_{9}^{13} \frac{1}{u} \cdot \frac{du}{2} = \frac{1}{2} [\ln |u|]_{9}^{13}$.
$I = \frac{1}{2} (\ln 13 - \ln 9) = \frac{1}{2} \ln \left( \frac{13}{9} \right)$.

$\int_{3}^{5} \frac{1}{2x + 3} dx$ is equal to

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