Question

6. Evaluate
integral of
dx/x-√x​

Answer 1

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\: Let \: A \: = \: \displaystyle \int \sf \: \dfrac{dx}{x - \sqrt{x} }$

Here, we use the method of Substitution,

$\rm :\longmapsto\: \red{ \rm \:Put \: \sqrt{x} = y}$

$\rm :\longmapsto\: \red{ \rm \: x = {y}^{2}}$

On differentiating w. r. t. x, we get

$\rm :\longmapsto\: \red{ \rm \: dx = {2ydy}}$

On substituting all these above values in given integral, we get

$\rm :\longmapsto\:A \: = \: \displaystyle \int \sf \: \dfrac{2y}{ {y}^{2} - y}dy$

$\: \sf \: \: = \: \: 2\displaystyle \int \sf \: \dfrac{1}{y - 1} dy$

$\: \sf \: \: = \: \: 2 \: log(y - 1) + c$

$\: \sf \: \: = \: \: 2 \: log( \sqrt{x} - 1) + c$

Additional Information :-

$\boxed{\red{\sf\:\displaystyle \int \sf \: sinx \: dx = - cosx + c}}$

$\boxed{\red{\sf\:\displaystyle \int \sf \: cosx \: dx = sinx + c}}$

$\boxed{\red{\sf\:\displaystyle \int \sf \: {sec}^{2} x \: dx = tanx + c}}$

$\boxed{\red{\sf\:\displaystyle \int \sf \: {cosec}^{2} x \: dx = - cotx + c}}$

$\boxed{\red{\sf\:\displaystyle \int \sf \: cosecx \: cotx \: dx = - cosecx + c}}$

$\boxed{\red{\sf\:\displaystyle \int \sf \: secx \: tanx \: dx = secx + c}}$

$\boxed{\red{\sf\:\displaystyle \int \sf \: tanx \: dx = log(secx)+ c}}$

$\boxed{\red{\sf\:\displaystyle \int \sf \: cotx \: dx = log(sinx)+ c}}$

$\boxed{\red{\sf\:\displaystyle \int \sf \: \dfrac{dx}{x} = log(x) + c}}$