Question

(√x+1/√x)³,Integrate the given function w.r.t. x considering them well defined and integrable over proper domain.

Answer 1

Dear Student,

Answer:$\int{({ \frac{ \sqrt{x} + 1 }{ \sqrt{x} } })^{3}} dx=x - \frac{2}{ \sqrt{x} } + 6 \sqrt{x} + 3 log(x ) + c$

Solution:

$\int{( { \frac{ \sqrt{x} + 1 }{ \sqrt{x} } )^{3}}} dx \\ \\ = \int{\frac{{( \sqrt{x} + 1 })^{3} }{x \sqrt{x} }} dx \\ \\ = \int{\frac{x \sqrt{x} + 1 + 3x + 3 \sqrt{x} }{x \sqrt{x}}} dx \\ \\ = \int{\frac{x \sqrt{x} }{x \sqrt{x}}} dx + \int{\frac{1}{x \sqrt{x}}} dx + \int{\frac{3}{ \sqrt{x}}} dx + \int{\frac{3}{x}}dx \\ \\ =\int{1}dx +\int{{x}^{ \frac{ - 3}{2}}} dx +\int{3 {x}^{ \frac{ - 1}{2}}} dx +\int{\frac{3}{x}} dx$

Now apply power rule,and do integration

$= x + \frac{ {x}^{ \frac{ - 3}{2} + 1} }{ \frac{ - 3}{2} + 1 } + 3 \frac{ {x}^{ - \frac{1}{2} + 1 } }{ - \frac{ 1}{2} + 1 } + 3 \: log \: x + c \\ \\ = x - \frac{2}{ \sqrt{x} } + 6 \sqrt{x} + 3 log(x ) + c$
Hope it helps you.