\( \int\left(1-x^2\right)^{-\frac{1}{2}} d x =\int \frac{1}{\left(1-x^2\right)^{1 / 2}} d x \)
\(=\int \frac{1}{\sqrt{1-x^2}} d x=\sin ^{-1} x+c\)

\(\int { \left( 1+{ x }^{ 2 } \right) ^{ -1 } } dx=\int { \frac { 1 }{ 1+{ x }^{ 2 } } } dx\)
= tan^-1 x + c

\(\int { { e }^{ x } } dx\) = e^x + c

Let I = \(\int { \sqrt { \left( x+1 \right) ^{ 2 }-4 } } \) dx = \(\int { \sqrt { \left( x+1 \right) ^{ 2 }-{ 2 }^{ 2 } } } \) dx
Let t = x + 1 \(\Rightarrow\) dt = dx
\(\therefore\)  I = \(\int { \sqrt { { t }^{ 2 }-{ 2 }^{ 2 } } } \) dt
Since \(\int { \sqrt { { x }^{ 2 }-{ a }^{ 2 } } } \) dx = \(\int { \sqrt { { x }^{ 2 }-{ a }^{ 2 } } } =\frac { x }{ 2 } \sqrt { { x }^{ 2 }-{ a }^{ 2 } } -\frac { { a }^{ 2 } }{ 2 } log\left| x+\sqrt { { x }^{ 2 }-{ a }^{ 2 } } \right| +c\)
we get I = \(\frac { t }{ 2 } \sqrt { { t }^{ 2 }-4 } -\frac { 4 }{ 2 } log\left| t+\sqrt { { r }^{ 2 }-4 } \right| +c\)
 I = \({x+1\over2}\sqrt{(x+1)^2-4}+2log|x+1+\sqrt{(x+1)^2}-4|+c\)

Let u = log x
Then e^u = x and
\(\frac{d u}{d x}=\frac{1}{x}\)
\(\therefore \frac{d u}{d x}=\frac{1}{e^u}\)
\(\therefore e^u d u=d x\)
\( \int \frac{\log x}{(1+\log x)^2} d x =\int \frac{u}{(1+u)^2} \times e^u d u\)
\(=\int e^u\left(\frac{(1+u)-1}{(1+u)^2}\right) d u \)
\(=\int e^u\left(\frac{1}{1+u}-\frac{1}{(1+u)^2}\right) d u \)
\(=e^u\left(\frac{1}{1+u}\right)+c \)
Putting u = log x and e^u = u
\(=x\left(\frac{1}{1+\log x}\right)+c \)
\(=\frac{x}{1+\log x}+c\)

\( \int \frac{1}{1+x^2} d x =\tan ^{-1} x+c\)
\( \therefore \int \frac{1}{1+36 x^2} d x =\int \frac{1}{1+(6 x)^2} d x \)
\(=\frac{\tan ^{-1} 6 x}{6}+c \)

\(\int { \frac { dx }{ \sqrt { 1-{ 81x }^{ 2 } } } } =\int { \frac { dx }{ \sqrt { 1-\left( 9x \right) ^{ 2 } } } } =\frac { sin^{ -1 }\left( 9x \right) }{ 9 } +c\)

\( \int \frac{1}{\sqrt{1-x^2}} d x=\sin ^{-1} x+c\)
\(\therefore \int \frac{1}{\sqrt{1-(4 x)^2}} d x=\frac{\sin ^{-1}(4 x)}{4}+c \)

\( \int \sec x \tan x d x=\sec x+c \)
\( \int \sec (2-15 x) \tan (2-15 x) d x \)
\(=\frac{\sec (2-15 x)}{-15}+c \)

\(\int \operatorname{cosec} x \cot x d x=\operatorname{-cosec} x+c\)
\(\therefore\int { cosec } \left( 5x+3 \right) cot(5x+3)dx\)
= \(-cosec\frac { \left( 5x+3 \right) }{ 5 } +c\)

\( \int \sec ^2 x d x=tan\ x+c\)
\( \therefore \int \sec ^2\left(\frac{x}{5}\right) d x=\frac{tan(x / 5)}{1 / 5}+c=5tan\left(\frac{x}{5}\right) +c \)

Let I =\(\int tan^{-1}({2x\over 1-x^2})dx\)
Putting x = \(tan \theta \Rightarrow dx=sec^2 \theta d \theta\)
Therefore, \(I=\int tan^{-1}({{2tan \theta}\over 1-tan^2 \theta})sec^2 \theta d\theta\)
\(=\int tan^{-1}(tan 2\theta)sec^2\theta d \theta\)
\(=\int 2\theta sec^2\theta d \theta\)
\(=2\int (\theta )(sec^2\theta d \theta)\)

Applying integration by parts
I = 2[\(\theta tan \theta- \int tan \theta d \theta\)]   \(tan \theta =x\)
= 2\((\theta tan \theta- log|sec \theta | )+c\)    \(sec \theta =\sqrt{1+x^2}\)
\(\int tan^{-1}({2x\over 1-x^2})dx=2x tan ^{-1}x-2log|\sqrt{1+x^2}|+c\)

\(\int x(1-x)^{17} d x =\int[(x-1)+1](1-x)^{17} d x \)
\(=\int[-(1-x)+1](1-x)^{17} d x \)
\(=\int\left[-1(1-x)^{16}+(1-x)^{17}\right] d x\)
\(=-\left(\frac{(1-x)^{19}}{19 \times(-1)}\right)+\frac{(1-x)^{18}}{18 x(-1)^{17}}+c\)
\(=\frac{(1-x)^{18}}{19}-\frac{(1-x)^{18}}{18}+c\)

Let A be the area of the wound after f days from the Sunday.
Then \(\frac{d A}{d t}=-\frac{6}{(t+2)^2} \text { (given) }\)
\(\therefore d A=-\frac{6}{(t+2)^2} d t \)
\(\therefore \int d A=\int \frac{-6}{(t+2)^2} d t\)
\( \therefore A =-6\left(\frac{-1}{t+2}\right)+c \)
\(=\frac{6}{t+2}+c \)
Given: \([A]_{t=2}=1.4\)
\( \therefore \frac{6}{2+2}+c =1.4 \)
\( 1.5+c =1.4 \)
\(\therefore c =-0.1\)
\(\therefore A =\frac{6}{t+2}-0.1\)
(i) \(\text { Area of the wound On sunday} =[A]_{t=1}\)
\( =\left[\frac{6}{t+2}-0.1\right]_{t=1}\)
\(=\frac{6}{3}-0.1 \)
\(=1.9 \mathrm{~cm}^2 \)
(ii) \(\text { Area of the wound On Thursday}=[A]_{t=5}\)
\(=\left[\frac{6}{t+2}-0.1\right]_{t=5} \)
\( =\frac{6}{5+2}-0.1 \)
\(=\frac{6}{7}-0.1\)
\(=0.76 \mathrm{~cm}^2 \)

\(=\int\left[\frac{6}{1+(3 x+2)^2}-\frac{12}{\sqrt{1-(3-4 x)^2}}\right] d x\)
\( =6 \int \frac{1}{1+(3 x+2)^2} d x-12 \int \frac{1}{\sqrt{1-(3-4 x)^2}} d x \)
\(=6\left(\frac{\operatorname{lan}^{-1}(3 x+2)}{3}\right)-12\left(\frac{\sin ^{-1}(3-4 x)}{-4}\right)+c \)
\(=2 \tan ^{-1}(3 x+2)+3 \sin ^{-1}(3-4 x)+c \)