Question

solve the question step by step​

Answer 1

Solution:

Given Integral:

$\displaystyle \rm \longrightarrow I = \int {x}^{ \frac{13}{2} }(1 + {x}^{ \frac{5}{2} } )^{ \frac{1}{2} } \: dx$

Let us assume that:

$\rm \longrightarrow u =1 + {x}^{ \frac{5}{2} }$

$\rm \longrightarrow \dfrac{du}{dx} = \dfrac{5{x}^{ \frac{3}{2} }}{2}$

$\rm \longrightarrow dx = \dfrac{2 \: du}{5{x}^{ \frac{3}{2} }}$

Now:

$\rm \longrightarrow u =1 + {x}^{ \frac{5}{2} }$

$\rm \longrightarrow u - 1 = {x}^{ \frac{5}{2} }$

$\rm \longrightarrow {x}^{5} = {(u- 1)}^{2}$

So, the integral changes to:

$\displaystyle \rm \longrightarrow I = \int {x}^{ \frac{13}{2} } \sqrt{u} \times\dfrac{2 \: du}{5 {x}^{ \frac{3}{2} } }$

$\displaystyle \rm \longrightarrow I = \dfrac{2}{5} \int {x}^{5} \sqrt{u} \: du$

$\displaystyle \rm \longrightarrow I = \dfrac{2}{5} \int {(u - 1)}^{2} \sqrt{u} \: du$

$\displaystyle \rm \longrightarrow I = \dfrac{2}{5} \int \bigg[ {u}^{ \frac{5}{2} } - 2 {u}^{ \frac{3}{2} } + \sqrt{u} \bigg] \: du$

$\displaystyle \rm \longrightarrow I = \dfrac{2}{5} \bigg[ \int{u}^{ \frac{5}{2} } \: du- \int2 {u}^{ \frac{3}{2} } \: du + \int\sqrt{u} \: du\bigg]$

$\displaystyle \rm \longrightarrow I = \dfrac{2}{5} \bigg[ \dfrac{2 {u}^{ \frac{7}{2} } }{7} - 2\int{u}^{ \frac{3}{2} } \: du + \int\sqrt{u} \: du\bigg]$

$\displaystyle \rm \longrightarrow I = \dfrac{2}{5} \bigg[ \dfrac{2 {u}^{ \frac{7}{2} } }{7} -2 \times \dfrac{2 {u}^{ \frac{5}{2} } }{5} + \dfrac{2 {u}^{ \frac{3}{2} } }{3} \bigg] + C$

$\displaystyle \rm \longrightarrow I = \dfrac{4 {u}^{ \frac{7}{2} } }{35} -\dfrac{8 {u}^{ \frac{5}{2} } }{25} + \dfrac{4 {u}^{ \frac{3}{2} } }{15} + C$

Substituting back the value of u, we get:

$\displaystyle \rm \longrightarrow I = \dfrac{4 {(1 + {x}^{ \frac{5}{2} } )}^{ \frac{7}{2} } }{35} -\dfrac{8 {(1 + {x}^{ \frac{5}{2} } )}^{ \frac{5}{2} } }{25} + \dfrac{4 {(1 + {x}^{ \frac{5}{2} } )}^{ \frac{3}{2} } }{15} + C$

★ Which is our required answer.

Additional Information:

$\boxed{\begin{array}{c|c}\bf f(x)&\bf\displaystyle\int\rm f(x)\:dx\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \rm k&\rm kx+C\\ \\ \rm sin(x)&\rm-cos(x)+C\\ \\ \rm cos(x)&\rm sin(x)+C\\ \\ \rm{sec}^{2}(x)&\rm tan(x)+C\\ \\ \rm{cosec}^{2}(x)&\rm-cot(x)+C\\ \\ \rm sec(x)\ tan(x)&\rm sec(x)+C\\ \\ \rm cosec(x)\ cot(x)&\rm-cosec(x)+C\\ \\ \rm tan(x)&\rm log(sec(x))+C\\ \\ \rm\dfrac{1}{x}&\rm log(x)+C\\ \\ \rm{e}^{x}&\rm{e}^{x}+C\\ \\ \rm x^{n},n\neq-1&\rm\dfrac{x^{n+1}}{n+1}+C\end{array}}$