Question

Integrate (√x+1/√x)^2 dx

Answer 1

Answer

• x²/2 + ln x + 2x + c

Given

$\bullet \;\;\; \rm \int \bigg(\sqrt{x}+\dfrac{1}{\sqrt{x}}\bigg)^2dx$

To Find

• Integral value

Concept

$\rm \bullet \;\;\; (a+b)^2=a^2+b^2+2ab\\\\ \rm \bullet \;\;\; \int x\ dx=\dfrac{x^2}{2}+c\\\\\rm \bullet \;\; \int \dfrac{dx}{x}=ln(x)\ +c\\\\\rm \bullet \;\;\; \int k\ dx=kx\ +c$

where ,

• k denotes constant value

Solution

$\rm \int \bigg( \sqrt{x}+\dfrac{1}{\sqrt{x}} \bigg)^2dx\\\\\implies \rm \int \bigg[(\sqrt{x})^2+\bigg(\dfrac{1}{\sqrt{x}}\bigg)^2 +2.\sqrt{x}.\dfrac{1}{\sqrt{x}} \bigg]dx\\\\\implies \rm \int [x+\dfrac{1}{x}+2]dx\\\\\implies \rm \dfrac{x^2}{2}+\ln x+2x+c$

Answer 2

Answer:

HELLO DEAR,

given function is ∫(√x - 1/√x)².dx

⇒ ∫(x + 1/x - 2).dx

⇒ ∫x.dx + ∫1/x.dx - 2∫dx

⇒x²/2 + log|x| - 2x + c

hence, the integration of [√x - 1/√x]² is (x²/2 + log|x| - 2x) + c.

I HOPE ITS HELP YOU