Question

evaluate the following integrals

Answer 1

Hey there!!

Here's your answer..

[tex] \int\ { ( \sqrt{x} + \frac{1}{ \sqrt{x} }) ^{2} } \, dx = \int\ {x + \frac{1}{x}+2 } \, dx [/tex]

But we know that $\int\ {x} \, dx = \frac{x^{2} }{2}$

Also, $\int\ { \frac{1}{x} } \, dx = log (x)$

And, $\int\ dx = x$

So, the required answer is $\frac{x ^{2} }{2} + log (x) + 2 x + C$

Hope it helps!!

Answer 2

Hey there!!

Here's your answer..

\int\ { ( \sqrt{x} + \frac{1}{ \sqrt{x} }) ^{2} } \, dx = \int\ {x + \frac{1}{x}+2 } \, dx∫ (x​+x​1​)2dx=∫ x+x1​+2dx 

But we know that \int\ {x} \, dx = \frac{x^{2} }{2}∫ xdx=2x2​ 

Also, \int\ { \frac{1}{x} } \, dx = log (x)∫ x1​dx=log(x) 

And, \int\ dx = x∫ dx=x 

So, the required answer is \frac{x ^{2} }{2} + log (x) + 2 x + C2x2​+log(x)+2x+C 

Hope it helps!!