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genyjoannerubiera

genyjoannerubiera

21.07.2019 •

Mathematics

Find the indefinite integral of $\int\limits {\frac{5}{x^\frac{1}{2}+x^\frac{3}{2}} \, dx$ i have been able to simplify it to $\int\limits {\frac{5\sqrt{x}}{x^3+x}} \, dx$ but that is confusing, i then did u-subsitution where $u=\sqrt{x}$ to obtain $\int\limits {\frac{5u}{u^6+u^2}} \, dx$ which simplified to $\int\limits {\frac{5}{u^5+u}} \, dx$ , a much nicer looking integrand however, i am still stuck ples show all work or be reported

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Ответ:

Yazminwilliams2504

14.03.2022

$\displaystyle10 \tan^{-1}( \sqrt {x}^{ } ) + \rm C$

Step-by-step explanation:

we would like to integrate the following integration:

$\displaystyle \int \frac{5}{ {x}^{ \frac{1}{2} } + {x}^{ \frac{3}{2} } } dx$

in order to do so we can consider using u-substitution

let our

$\displaystyle u = {x}^{ \frac{1}{2} } \quad \text{and} \quad du = \frac{ {x}^{ - \frac{1}{2} } }{2}$

to apply substitution we need a little bit arrangement

multiply both Integral and integrand by 2 and ½

$\displaystyle 2\int \frac{1}{2} \cdot\frac{5}{ {x}^{ \frac{1}{2} } + {x}^{ \frac{3}{2} } } dx$

factor out $x^{\frac{1}{2}}$ :

$\displaystyle 2\int \frac{1}{2 {x}^{ \frac{1}{2} } } \cdot\frac{5}{ (1+ {x}^{ } )} dx$

recall law of exponent:

$\displaystyle 2\int \frac{ {x}^{ - \frac{1}{2} } }{2 } \cdot\frac{5}{ (1+ {x}^{ } )} dx$

apply substitution:

$\displaystyle 2\int \frac{5}{ 1+ {x}^{ } } du$

rewrite x as $\big(x^{\frac{1}{2}}\big)^{2}$ :

$\displaystyle 2\int \frac{5}{ 1+ ( {x ^{ \frac{1}{2} } })^{ 2 } } du$

substitute:

$\displaystyle 2\int \frac{5}{ 1+ ( u)^{ 2 } } du$

recall integration rule of inverse trig:

$\displaystyle 2 \times 5 \tan^{-1}(u)$

simplify multiplication:

$\displaystyle10 \tan^{-1}(u)$

substitute back:

$\displaystyle10 \tan^{-1}( {x}^{ \frac{1}{2} } )$

simplify if needed:

$\displaystyle10 \tan^{-1}( \sqrt{x} )$

finally we of course have to add constant of integration:

$\displaystyle10 \tan^{-1}( \sqrt {x}^{ } ) + \rm C$

and we are done!

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Ответ:

001054213

14.03.2022

The easiest way to calculate this integral is substitution.

$$\int\dfrac{5}{x^\frac{1}{2}+x^\frac{3}{2}}\,dx=5\int\dfrac{1}{x^\frac{1}{2}+(x^\frac{1}{2})^3}\,dx=5\int\dfrac{1}{\sqrt{x}+(\sqrt{x})^3}\,dx=(\star)$

Now we can substitute $u=\sqrt{x}$ and then:

$du=\dfrac{1}{2\sqrt{x}}\,dx\qquad\implies\qquad dx=2\sqrt{x}\,du=2u\,du$

So:

$$(\star)=5\int\dfrac{1}{\sqrt{x}+(\sqrt{x})^3}\,dx=5\int\dfrac{2u}{u+u^3}\,du=10\int\dfrac{1}{1+u^2}\,dx=$

$=10\arctan(u)+C=\boxed{10\arctan(\sqrt{x})+C}$

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Ответ:

dranio

27.10.2020

0.6 words per minute

Step-by-step explanation: Hopefully this helps! :)

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