Answer:
[tex] \displaystyle10 \tan^{-1}( \sqrt {x}^{ } ) + \rm C[/tex]
Step-by-step explanation:
we would like to integrate the following integration:
[tex] \displaystyle \int \frac{5}{ {x}^{ \frac{1}{2} } + {x}^{ \frac{3}{2} } } dx[/tex]
in order to do so we can consider using u-substitution
let our
[tex] \displaystyle u = {x}^{ \frac{1}{2} } \quad \text{and} \quad du = \frac{ {x}^{ - \frac{1}{2} } }{2} [/tex]
to apply substitution we need a little bit arrangement
multiply both Integral and integrand by 2 and ½
[tex]\displaystyle 2\int \frac{1}{2} \cdot\frac{5}{ {x}^{ \frac{1}{2} } + {x}^{ \frac{3}{2} } } dx[/tex]
factor out [tex]x^{\frac{1}{2}}[/tex]:
[tex]\displaystyle 2\int \frac{1}{2 {x}^{ \frac{1}{2} } } \cdot\frac{5}{ (1+ {x}^{ } )} dx[/tex]
recall law of exponent:
[tex]\displaystyle 2\int \frac{ {x}^{ - \frac{1}{2} } }{2 } \cdot\frac{5}{ (1+ {x}^{ } )} dx[/tex]
apply substitution:
[tex]\displaystyle 2\int \frac{5}{ 1+ {x}^{ } } du[/tex]
rewrite x as [tex]\big(x^{\frac{1}{2}}\big)^{2}[/tex]:
[tex]\displaystyle 2\int \frac{5}{ 1+ ( {x ^{ \frac{1}{2} } })^{ 2 } } du[/tex]
substitute:
[tex]\displaystyle 2\int \frac{5}{ 1+ ( u)^{ 2 } } du[/tex]
recall integration rule of inverse trig:
[tex] \displaystyle 2 \times 5 \tan^{-1}(u)[/tex]
simplify multiplication:
[tex] \displaystyle10 \tan^{-1}(u)[/tex]
substitute back:
[tex] \displaystyle10 \tan^{-1}( {x}^{ \frac{1}{2} } )[/tex]
simplify if needed:
[tex] \displaystyle10 \tan^{-1}( \sqrt{x} )[/tex]
finally we of course have to add constant of integration:
[tex] \displaystyle10 \tan^{-1}( \sqrt {x}^{ } ) + \rm C[/tex]
and we are done!
