Question

integrate (√x + 1/√x)^2​

Answer 1

Answer:

$\mathrm{\dfrac{x^2}{2} + 2x + log|x| + c}$

Step-by-step explanation:

Given :

$\mathrm{\displaystyle \int \bigg(\sqrt x + \dfrac{1}{\sqrt{x} }\bigg)^2 \, dx}$

To find :

Evaluate the given integral

Solution :

$\longmapsto \mathrm{\displaystyle \int \bigg(\sqrt x + \dfrac{1}{\sqrt{x} }\bigg)^2 \, dx}$

We know that,

$\boxed{\mathrm{(a+b)^2 = a^2 + 2ab + b^2}}$

$\implies \mathrm{\displaystyle \int \Bigg((\sqrt x)^2 + 2\cancel{\sqrt x}.\dfrac{1}{\cancel{\sqrt x}} + \bigg(\dfrac{1}{\sqrt{x} }\bigg)^2 \Bigg) \, dx}$

$\implies\mathrm{\displaystyle \int \bigg(x + 2 + \dfrac{1}{x}\bigg) \, dx}$

We know that,

$\boxed{\begin{array}{cc} \mathrm{\displaystyle \int x^n \, dx = \dfrac{x^{n+1}}{n+1} + c \ \ (n \neq -1)}\\ \\ \mathrm{\displaystyle \int k \, dx = kx + c} \\ \\ \mathrm{\displaystyle \int \dfrac{1}{x} \, dx = log|x| + c} \end{array}}$

So, applying these, we get,

$\implies \mathrm{\dfrac{x^2}{2} + 2x + log|x| + c}$

$\therefore \underline{\boxed{\mathbf{\displaystyle \int \bigg(\sqrt x + \dfrac{1}{\sqrt{x} }\bigg)^2 \, dx} = \mathbf{\dfrac{x^2}{2} + 2x + log|x| + c} }}$

❖ Extra information ;

⟡ Some basic integrals

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\\\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx+c \\\\ \sf x^n \ (n \neq -1)& \sf \dfrac{x^{n+1}}{n+1} + c \\\\ \sf \dfrac{1}{x} & \sf logx+ c\\\\ \sf {e}^{x} & \sf {e}^{x}+c\\\\ \sf sinx & \sf - \: cosx+ c \\\\ \sf cosx & \sf \: sinx + c\\\\ \sf {sec}^{2} x & \sf tanx + c\\\\ \sf {cosec}^{2}x & \sf - cotx+ c \\\\ \sf secx \: tanx & \sf secx + c\\\\ \sf cosecx \: cotx& \sf -\: cosecx + c\end{array}} \\ \end{gathered}$

Hope it helps!!