Question

refer the attachment

Answer 1

Question :-

Evaluate the following integral :-

$\rm :\longmapsto\:\displaystyle\int\rm \frac{1}{x + \sqrt{x - 1} } \: dx$

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \frac{1}{x + \sqrt{x - 1} } \: dx$

To solve this integral, we use method of Substitution

Let substitute

$\red{\rm :\longmapsto\: \sqrt{x - 1} = y}$

On squaring both sides, we get

$\red{\rm :\longmapsto\: x - 1 = {y}^{2} }$

$\red{\rm :\longmapsto\: x = {y}^{2} + 1}$

So,

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

Now, Substituting all these values in given integral, we get

$\rm \: = \:\displaystyle\int\rm \frac{2y}{{y}^{2} + 1 + y} \: dy$

can be rewritten as

$\rm \: = \:\displaystyle\int\rm \frac{2y}{{y}^{2}+ y + 1} \: dy$

$\rm \: = \:\displaystyle\int\rm \frac{2y + 1 - 1}{{y}^{2}+ y + 1} \: dy$

$\rm \: = \:\displaystyle\int\rm \frac{2y + 1}{{y}^{2}+ y + 1} \: dy - \displaystyle\int\rm \frac{1}{ {y}^{2} + y + 1}dy$

$\rm \:=\:\displaystyle\int\rm \frac{\dfrac{d}{dy} ( {y}^{2} + y + 1)}{{y}^{2}+ y + 1} \: dy - \displaystyle\int\rm \frac{1}{ {y}^{2}+ y +\dfrac{1}{4} - \dfrac{1}{4}+ 1}dy$

We know,

$\boxed{ \tt{ \: \displaystyle\int\rm \frac{f'(x)}{f(x)} \: dx \: = \: logf(x) + c \: }}$

Now, using this formula in first integral we get

$\rm \: = \:log | {y}^{2} + y + 1 | - \displaystyle\int\rm \frac{1}{\bigg[ {y}^{2} + 2 \times \dfrac{1}{2} \times y + \dfrac{1}{4} \bigg] + \dfrac{3}{4} }$

$\rm \: = \:log | {y}^{2} + y + 1| - \displaystyle\int\rm \frac{1}{ {\bigg[y + \dfrac{1}{2} \bigg]}^{2} + {\bigg[\dfrac{ \sqrt{3} }{2} \bigg]}^{2} }$

We know,

$\boxed{ \tt{ \: \displaystyle\int\rm \frac{dx}{ {x}^{2} + {a}^{2}} = \frac{1}{a} {tan}^{ - 1} \frac{x}{a} + c }}$

So, using this identity, we get

$\rm \: = \:log | {y}^{2} + y + 1| - \dfrac{1}{\dfrac{ \sqrt{3} }{2} } {tan}^{ - 1}\bigg[\dfrac{y + \dfrac{1}{2} }{\dfrac{ \sqrt{3} }{2} } \bigg] + c$

$\rm \: = \:log | {y}^{2} + y + 1| - \dfrac{2}{ \sqrt{3} } {tan}^{ - 1}\bigg[\dfrac{2y + 1}{ \sqrt{3} } \bigg] + c$

On substituting the value of y, we get

$\rm \: = \:log | x - 1 + \sqrt{x + 1} + 1| - \dfrac{2}{ \sqrt{3} } {tan}^{ - 1}\bigg[\dfrac{2 \sqrt{x + 1} + 1}{ \sqrt{3} } \bigg] + c$

$\rm \: = \:log | x+ \sqrt{x + 1}| - \dfrac{2}{ \sqrt{3} } {tan}^{ - 1}\bigg[\dfrac{2 \sqrt{x + 1} + 1}{ \sqrt{3} } \bigg] + c$

More to know :-

$\boxed{ \tt{ \: \displaystyle\int\rm \frac{dx}{ {x}^{2} - {a}^{2} } = \frac{1}{2a}log\bigg | \frac{x - a}{x + a} \bigg| + c}}$

$\boxed{ \tt{ \: \displaystyle\int\rm \frac{dx}{ \sqrt{ {a}^{2} - {x}^{2} } } = {sin}^{ - 1} \frac{x}{a} + c}}$

$\boxed{ \tt{ \: \displaystyle\int\rm \frac{dx}{ \sqrt{ {x}^{2} - {a}^{2} } } = log \bigg|x + \sqrt{ {x}^{2} - {a}^{2} }\bigg | + c}}$

$\boxed{ \tt{ \: \displaystyle\int\rm \frac{dx}{ \sqrt{ {x}^{2} + {a}^{2} } } = log \bigg|x + \sqrt{ {x}^{2} + {a}^{2} }\bigg | + c}}$

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

Answer 2

Answer:

Question :-

Evaluate the following integral :-

$\rm :\longmapsto\:\displaystyle\int\rm \frac{1}{x + \sqrt{x - 1} } \: dx$

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \frac{1}{x + \sqrt{x - 1} } \: dx$

To solve this integral, we use method of Substitution

Let substitute

$\red{\rm :\longmapsto\: \sqrt{x - 1} = y}$

On squaring both sides, we get

$\red{\rm :\longmapsto\: x - 1 = {y}^{2} }$

$\red{\rm :\longmapsto\: x = {y}^{2} + 1}$

So,

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

Now, Substituting all these values in given integral, we get

$\rm \: = \:\displaystyle\int\rm \frac{2y}{{y}^{2} + 1 + y} \: dy$

can be rewritten as

$\rm \: = \:\displaystyle\int\rm \frac{2y}{{y}^{2}+ y + 1} \: dy$

$\rm \: = \:\displaystyle\int\rm \frac{2y + 1 - 1}{{y}^{2}+ y + 1} \: dy$

$\rm \: = \:\displaystyle\int\rm \frac{2y + 1}{{y}^{2}+ y + 1} \: dy - \displaystyle\int\rm \frac{1}{ {y}^{2} + y + 1}dy$

$\rm \:=\:\displaystyle\int\rm \frac{\dfrac{d}{dy} ( {y}^{2} + y + 1)}{{y}^{2}+ y + 1} \: dy - \displaystyle\int\rm \frac{1}{ {y}^{2}+ y +\dfrac{1}{4} - \dfrac{1}{4}+ 1}dy$

We know,

$\boxed{ \tt{ \: \displaystyle\int\rm \frac{f'(x)}{f(x)} \: dx \: = \: logf(x) + c \: }}$

Now, using this formula in first integral we get

$\rm \: = \:log | {y}^{2} + y + 1 | - \displaystyle\int\rm \frac{1}{\bigg[ {y}^{2} + 2 \times \dfrac{1}{2} \times y + \dfrac{1}{4} \bigg] + \dfrac{3}{4} }$

$\rm \: = \:log | {y}^{2} + y + 1| - \displaystyle\int\rm \frac{1}{ {\bigg[y + \dfrac{1}{2} \bigg]}^{2} + {\bigg[\dfrac{ \sqrt{3} }{2} \bigg]}^{2} }$

We know,

$\boxed{ \tt{ \: \displaystyle\int\rm \frac{dx}{ {x}^{2} + {a}^{2}} = \frac{1}{a} {tan}^{ - 1} \frac{x}{a} + c }}$

So, using this identity, we get

$\rm \: = \:log | {y}^{2} + y + 1| - \dfrac{1}{\dfrac{ \sqrt{3} }{2} } {tan}^{ - 1}\bigg[\dfrac{y + \dfrac{1}{2} }{\dfrac{ \sqrt{3} }{2} } \bigg] + c$

$\rm \: = \:log | {y}^{2} + y + 1| - \dfrac{2}{ \sqrt{3} } {tan}^{ - 1}\bigg[\dfrac{2y + 1}{ \sqrt{3} } \bigg] + c$

On substituting the value of y, we get

$\rm \: = \:log | x - 1 + \sqrt{x + 1} + 1| - \dfrac{2}{ \sqrt{3} } {tan}^{ - 1}\bigg[\dfrac{2 \sqrt{x + 1} + 1}{ \sqrt{3} } \bigg] + c$

$\rm \: = \:log | x+ \sqrt{x + 1}| - \dfrac{2}{ \sqrt{3} } {tan}^{ - 1}\bigg[\dfrac{2 \sqrt{x + 1} + 1}{ \sqrt{3} } \bigg] + c$

More to know :-

$\boxed{ \tt{ \: \displaystyle\int\rm \frac{dx}{ {x}^{2} - {a}^{2} } = \frac{1}{2a}log\bigg | \frac{x - a}{x + a} \bigg| + c}}$

$\boxed{ \tt{ \: \displaystyle\int\rm \frac{dx}{ \sqrt{ {a}^{2} - {x}^{2} } } = {sin}^{ - 1} \frac{x}{a} + c}}$

$\boxed{ \tt{ \: \displaystyle\int\rm \frac{dx}{ \sqrt{ {x}^{2} - {a}^{2} } } = log \bigg|x + \sqrt{ {x}^{2} - {a}^{2} }\bigg | + c}}$

$\boxed{ \tt{ \: \displaystyle\int\rm \frac{dx}{ \sqrt{ {x}^{2} + {a}^{2} } } = log \bigg|x + \sqrt{ {x}^{2} + {a}^{2} }\bigg | + c}}$

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