Check the solution to "Evaluate the integrals ∫0ln⁡5er(3er+1)−32dr"

shadsiei

Answered question

2021-10-08

Evaluate the integrals

\int_{0}^{\ln 5} e^{r} {(3 e^{r} + 1)}^{- \frac{3}{2}} d r

Cullen

Skilled2021-10-09Added 89 answers

Given:

\int_{0}^{\ln (5)} e^{r} {(3 e^{r} + 1)}^{\frac{- 3}{2}} d r

Calculation:

\int_{0}^{\ln (5)} e^{r} {(3 e^{r} + 1)}^{\frac{- 3}{2}} d r

= \int_{0}^{\ln (5)} \frac{e^{r}}{{(3 e^{r} + 1)}^{\frac{3}{2}}} d r

put

3 e^{r} + 1 = u \Rightarrow 3 e^{r} d r = d u

when

r = 0 \Rightarrow 3 e^{0} + 1 = u \Rightarrow 3 + 1 = u \Rightarrow u = 4

when

r = \ln 5 \Rightarrow 3 e^{\ln 5} + 1 = u \Rightarrow 3 \times 5 + 1 = u \Rightarrow 16

\int_{4}^{16} \frac{1}{3 u^{\frac{3}{2}}} d u

= \frac{1}{3} \int_{4}^{16} u^{\frac{- 3}{2}} d u

= \frac{1}{3} {[\frac{u^{\frac{- 3}{2} + 1}}{\frac{- 3}{2} + 1}]}_{4}^{16}

= \frac{1}{3} {[\frac{u^{\frac{- 1}{2}}}{\frac{- 1}{2}}]}_{4}^{16}

= \frac{- 2}{3} {[\frac{1}{u^{\frac{1}{2}}}]}_{4}^{16}

= \frac{- 2}{3} [\frac{1}{{(16)}^{\frac{1}{2}}} - \frac{1}{{(4)}^{\frac{1}{2}}}]

= \frac{- 2}{3} [\frac{1}{4} - \frac{1}{2}]

= \frac{- 2}{3} \times \frac{- 1}{4}

= \frac{1}{6}

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?