\(\frac{1}{2}\left[x \cos ^{-1} x-\sqrt{1-x^2}\right]+C\)

Let  \(\frac { dx }{ { x }^{ 2 }(x+1) } \equiv \ \frac { A }{ x } +\frac { B }{ { x }^{ 2 } } +\frac { C }{ 1+C } \)
\(\Rightarrow 1\equiv \ Ax(1+x)+B(1+x)+C{ x }^{ 2 }.\)
Putting   \(x=0, 1=B \Rightarrow B=1.\)
Putting  \(x=-1, 1=C{ (-1) }^{ 2 }\Rightarrow C=1\)
Comparing coeffs. of  \({ x }^{ 2 },0=A+C\)
\(\Rightarrow A=-C=-1\)
ஃ From(1), \(\frac { 1 }{ { x }^{ 2 }(1+x) } =\frac { -1 }{ x } +\frac { 1 }{ { x }^{ 2 } } +\frac { 1 }{ (1+x) } .\)
\(\therefore \ \int _{ 1 }^{ 3 }{ \frac { dx }{ { x }^{ 2 }(1+x) } } =\int _{ 1 }^{ 3 }{ \left( -\frac { 1 }{ x } +\frac { 1 }{ { x }^{ 2 } } +\frac { 1 }{ (1+x) } \right) } dx.\)
\(=-\int _{ 1 }^{ 3 }{ \frac { 1 }{ x } } dx+\int _{ 1 }^{ 3 }{ { x }^{ -2 } } dx+\int _{ 1 }^{ 3 }{ \frac { 1 }{ 1+x } } dx\)
\(=-{ \left[ \log { \left| x \right| } \right] }_{ 1 }^{ 3 }+{ \left[ \frac { { x }^{ -1 } }{ -1 } \right] }_{ 1 }^{ 3 }+{ \left[ \log { \left| 1+x \right| } \right] }_{ 1 }^{ 3 }\)
\(=-\left[ \log { \left| 3 \right| } -\log { \left| 1 \right| } \right] -{ \left[ \frac { 1 }{ x } \right] }_{ 1 }^{ 3 }+\left[ \log { \left| 1+3 \right| } -\log { \left| 2 \right| } \right] \)
\(=-\left[ \log { 3 } -\log { 1 } \right] -\left[ \frac { 1 }{ 2 } -1 \right] +\left[ \log { 4 } -\log { 2 } \right] =-\left[ \log { 3 } \right] +\frac { 2 }{ 3 } +\log { \frac { 4 }{ 2 } } \)
\(=-\log { 3 } +\frac { 2 }{ 3 } +\log { 2 } =\log { \frac { 2 }{ 3 } } +\frac { 2 }{ 3 } \)
\(=\frac { 2 }{ 3 } +\log { \frac { 2 }{ 3 } } .\)

\(Put\ { x }^{ 2 }=t\) so that  \(i.e. xdx=\frac { 1 }{ 2 } dt.\)
When x=0,t=0. when x=1,t=1.
\(\therefore \int _{ 0 }^{ 1 }{ x } { e }^{ { x }^{ 2 } }dx=\int _{ 0 }^{ 1 }{ { e }^{ t }\frac { dt }{ 2 } } =\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ { e }^{ t } } dt\)
\( =\frac { 1 }{ 2 } { \left[ { e }^{ t } \right] }_{ 0 }^{ 1 }\)
\(=\frac { 1 }{ 2 } \left[ { e }^{ 1 }-{ e }^{ 0 } \right] =\frac { 1 }{ 2 } (e-1).\)

\(I= \int { \frac { 3x }{ 1+{ 2x }^{ 4 } } } dx\)
Put  \({ x }^{ 2 }=t\) so that  \(2x\ dx=dt\)
\( i.e. x\ dx=\frac { 1 }{ 2 } dt.\)
\(\therefore \ I=\int { \frac { 3.\frac { 1 }{ 2 } dt }{ 1+{ 2x }^{ 4 } } } =\frac { 3 }{ 4 } \int { \frac { dt }{ \frac { 1 }{ 2 } +{ t }^{ 2 } } } \)
\( =\frac { 3 }{ 4 } \int { \frac { dt }{ { \left( \sqrt { \frac { 1 }{ 2 } } \right) }^{ 2 }+{ t }^{ 2 } } } \ \left[ From:\ \int { \frac { 1 }{ { a }^{ 2 }+{ x }^{ 2 } } } dx \right] \)
\(=\frac { 3 }{ 4 } .\frac { 1 }{ \sqrt { \frac { 1 }{ 2 } } } \tan ^{ -1 }{ \frac { t }{ \sqrt { \frac { 1 }{ 2 } } } } +C\)
\(=\frac { 3 }{ 2\sqrt { 2 } } \tan ^{ -1 }{ (\sqrt { { 2x }^{ 2 } } ) } +C\)

\(I= \int { \sin { 3x } \cos { 4x } } dx\)
\(=\frac { 1 }{ 2 } \int { \left( \sin { (3x+4x)+\sin { (3x-4x) } } \right) } dx\)
\(=\frac { 1 }{ 2 } \int { (\sin { 7x } -\sin { x } ) } dx\)
\(=\frac { 1 }{ 2 } \int { \sin { 7x } } dx-\frac { 1 }{ 2 } \left( \sin { x } dx \right) \)
\(=\frac { 1 }{ 2 } \left( -\frac { \cos { 7x } }{ 7 } \right) -\frac { 1 }{ 2 } \left( -\cos { x } \right) +C\)
\(=-\frac { 1 }{ 14 } \cos { 7x } +\frac { 1 }{ 2 } \cos { x } +C\)

\(I= \int { \frac { \cos { \sqrt { x } } }{ \sqrt { x } } } dx.\)
Put \(\sqrt { x } =t\) so that \(\frac { 1 }{ 2\sqrt { x } } dx=dt\)
\(i.e. \frac { 1 }{ \sqrt { x } } dx=2dt.\)
\( \therefore I= \int { \cos { t } } .2dt=2\int { \cos { t } } dt\)
\(=2 \sin { t } +C=2 \sin { \sqrt { c } } +C\)

\(I=\quad \int { \frac { { e }^{ 2x }-{ e }^{ -2x } }{ { e }^{ 2x }+{ e }^{ -2x } } } dx.\)
Put \({ e }^{ 2x }+{ e }^{ -2x }=t\) so that  \(2({ e }^{ 2x }-{ e }^{ -2x })\ dx=dt\)
\(i.e.\ ({ e }^{ 2x }-{ e }^{ -2x })dx=\frac { 1 }{ 2 } dt\)
\(\therefore I= \int { \frac { \frac { 1 }{ 2 } dt }{ t } = } \frac { 1 }{ 2 } \int { \frac { dt }{ 2 } } =\frac { 1 }{ 2 } \log { \left| t \right| } +C\)
\(=\frac { 1 }{ 2 } \log { \left| { e }^{ 2x }+{ e }^{ -2x } \right| } +C.\)

\(\text { Let } \tan ^{-1} x=t \)
\(\therefore \frac{1}{1+x^{2}} d x=d t \)
\(\Rightarrow \int \frac{e^{\tan ^{-1} x}}{1+x^{2}} d x=\int e^{t} d t \)
\(=e^{t}+\mathrm{C} \)
\(=e^{\tan ^{-1} x}+\mathrm{C} \)

\(I=\int { \frac { x }{ \sqrt { x+4 } } } dx=\int { \frac { (x+4)-4 }{ \sqrt { x+4 } } } dx\)
\( =\int { \sqrt { x+4 } } dx-4\int { \frac { x }{ \sqrt { x+4 } } } dx\)
\(=\int { { (x+4) }^{ \frac { 1 }{ 2 } } } dx-4\int { { (x+4) }^{ -\frac { 1 }{ 2 } } } dx\)
\(=\frac { { (x+4) }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } -4\frac { { (x+4) }^{ \frac { 1 }{ 2 } } }{ \frac { 1 }{ 2 } } +C\)
\(=\frac { 2 }{ 3 } { (x+4) }^{ \frac { 3 }{ 2 } }-8{ (x+4) }^{ \frac { 1 }{ 2 } }+C\)

Let log x=t so that 
\(\frac { 1 }{ x } dx=dt\)
\(\therefore I= \int { { t }^{ 2 } } dt=\frac { { t }^{ 3 } }{ 3 } +C\)
\(=\frac { { \left( \log { x } \right) }^{ 3 } }{ x } +C\)

(i) Using the formula, we express
\(x+2=\mathrm{A} \frac{d}{d x}\left(2 x^{2}+6 x+5\right)+\mathrm{B}=\mathrm{A}(4 x+6)+\mathrm{B}\)
Equating the coefficients of x and the constant terms from both sides, we get
\(4 \mathrm{~A}=1 \text { and } 6 \mathrm{~A}+\mathrm{B}=2 \text { or } \mathrm{A}=\frac{1}{4} \text { and } \mathrm{B}=\frac{1}{2}\)
\(\text {Therefore, }\int \frac{x+2}{2 x^{2}+6 x+5}=\frac{1}{4} \int \frac{4 x+6}{2 x^{2}+6 x+5} d x+\frac{1}{2} \int \frac{d x}{2 x^{2}+6 x+5}\)
\(=\frac{1}{4} \mathrm{I}_{1}+\frac{1}{2} \mathrm{I}_{2} \quad \text { (say) } \)
In I₁, put 2x² + 6x + 5 = t, so that (4x + 6) dx = dt
\(\text {Therefore, } \ \mathrm{I}_{1}=\int \frac{d t}{t}=\log |t|+\mathrm{C}_{1}\)
= log | 2x² + 6x + 5 | + C₁
\(\text {and } I_{2}=\int \frac{d x}{2 x^{2}+6 x+5}=\frac{1}{2} \int \frac{d x}{x^{2}+3 x+\frac{5}{2}} \)
\(=\frac{1}{2} \int \frac{d x}{\left(x+\frac{3}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}} \)
\(\text {Put } x+\frac{3}{2}=t, \text { so that } d x=d t, \text { we get }\)
\(\mathrm{I}_{2} =\frac{1}{2} \int \frac{d t}{t^{2}+\left(\frac{1}{2}\right)^{2}}=\frac{1}{2 \times \frac{1}{2}} \tan ^{-1} 2 t+\mathrm{C}_{2} \)
\(=\tan ^{-1} 2\left(x+\frac{3}{2}\right)+\mathrm{C}_{2}=\tan ^{-1}(2 x+3)+\mathrm{C}_{2} \)
Using (2) and (3) in (1), we get
\(\int \frac{x+2}{2 x^{2}+6 x+5} d x=\frac{1}{4} \log \left|2 x^{2}+6 x+5\right|+\frac{1}{2} \tan ^{-1}(2 x+3)+C\)
\(\text {where, }\mathrm{C}=\frac{\mathrm{C}_{1}}{4}+\frac{\mathrm{C}_{2}}{2} \)
(ii) This integral is of the form given. Let us express
\(x+3=A \frac{d}{d x}\left(5-4 x-x^{2}\right)+B=A(-4-2 x)+B\)
Equating the coefficients of x and the constant terms from both sides, we get
\(-2 \mathrm{~A}=1 \text { and }-4 \mathrm{~A}+\mathrm{B}=3, \text { i.e., } \mathrm{A}=-\frac{1}{2} \text { and } \mathrm{B}=1\)
\(\text {Therefore, } \int \frac{x+3}{\sqrt{5-4 x-x^{2}}} d x=-\frac{1}{2} \int \frac{(-4-2 x) d x}{\sqrt{5-4 x-x^{2}}}+\int \frac{d x}{\sqrt{5-4 x-x^{2}}}\)
\(=-\frac{1}{2} \mathrm{I}_{1}+\mathrm{I}_{2}\)
In I₁, put 5 – 4x – x² = t, so that (– 4 – 2x) dx = dt.
\(\text {Therefore, } \quad \mathrm{I}_{1}=\int \frac{(-4-2 x) d x}{\sqrt{5-4 x-x^{2}}}=\int \frac{d t}{\sqrt{t}}=2 \sqrt{t}+\mathrm{C}_{1}\)
\(=2 \sqrt{5-4 x-x^{2}}+C_{1}\)
\(\text {Now consider } \quad \mathrm{I}_{2}=\int \frac{d x}{\sqrt{5-4 x-x^{2}}}=\int \frac{d x}{\sqrt{9-(x+2)^{2}}}\)
Put x+2=t , so that dx=dt . 
\(\text {Therefore, } \quad \mathrm{I}_{2}=\int \frac{d t}{\sqrt{3^{2}-t^{2}}}=\sin ^{-1} \frac{t}{3}+\mathrm{C}_{2}\)
\(=\sin ^{-1} \frac{x+2}{3}+C_{2}\)
Substituting (2) and (3) in (1) , we obtain  \(\)
\(\int \frac{x+3}{\sqrt{5-4 x-x^{2}}}=-\sqrt{5-4 x-x^{2}}+\sin ^{-1} \frac{x+2}{3}+\mathrm{C}, \text { where } \mathrm{C}=\mathrm{C}_{2}-\frac{\mathrm{C}_{1}}{2}\)

(i) We have x² – 6x + 13 = x ²– 6x + 3² – 3² + 13 = (x – 3)² + 4
\(\text { So, } \ \int \frac{d x}{x^{2}-6 x+13}=\int \frac{1}{(x-3)^{2}+2^{2}} d x\)
Let x – 3 = t. Then dx = dt
\(\int \frac{d x}{x^{2}-6 x+13}=\int \frac{d t}{t^{2}+2^{2}}=\frac{1}{2} \tan ^{-1} \frac{t}{2}+C\)
\(=\frac{1}{2} \tan ^{-1} \frac{x-3}{2}+\mathrm{C}\)
(ii) The given integral is of the form 7.4 (7). We write the denominator of the integrand,
\(3 x^{2}+13 x-10=3\left(x^{2}+\frac{13 x}{3}-\frac{10}{3}\right)\)
\(=3\left[\left(x+\frac{13}{6}\right)^{2}-\left(\frac{17}{6}\right)^{2}\right] \text { (completing the square) }\)
\(\text { Thus } \int \frac{d x}{3 x^{2}+13 x-10}=\frac{1}{3} \int \frac{d x}{\left(x+\frac{13}{6}\right)^{2}-\left(\frac{17}{6}\right)^{2}} \)
\(\text { Put } x+\frac{13}{6}=t \text { . Then } d x=d t . \)
\(\text { Therefore, } \quad \int \frac{d x}{3 x^{2}+13 x-10}=\frac{1}{3} \int \frac{d t}{t^{2}-\left(\frac{17}{6}\right)^{2}} \)
\(=\frac{1}{3 \times 2 \times \frac{17}{6}} \log \left|\frac{t-\frac{17}{6}}{t+\frac{17}{6}}\right|+C_{1} \)
\(=\frac{1}{17} \log \left|\frac{x+\frac{13}{6}-\frac{17}{6}}{x+\frac{13}{6}+\frac{17}{6}}\right|+C_{1} \)
\(=\frac{1}{17} \log \left|\frac{6 x-4}{6 x+30}\right|+C_{1} \)
\(=\frac{1}{17} \log \left|\frac{3 x-2}{x+5}\right|+C_{1}+\frac{1}{17} \log \frac{1}{3} \)
\(=\frac{1}{17} \log \left|\frac{3 x-2}{x+5}\right|+C, \text { where } C=C_{1}+\frac{1}{17} \log \frac{1}{3} \)
\(\text { (iii) We have } \int \frac{d x}{\sqrt{5 x^{2}-2 x}}=\int \frac{d x}{\sqrt{5\left(x^{2}-\frac{2 x}{5}\right)}}\)
\(=\frac{1}{\sqrt{5}} \int \frac{d x}{\sqrt{\left(x-\frac{1}{5}\right)^{2}-\left(\frac{1}{5}\right)^{2}}} \text { (completing the square) }\)
\(\text { Put } x-\frac{1}{5}=t . \text { Then } d x=d t \text { . }\)
\(\text { Therefore, } \quad \int \frac{d x}{\sqrt{5 x^{2}-2 x}}=\frac{1}{\sqrt{5}} \int \frac{d t}{\sqrt{t^{2}-\left(\frac{1}{5}\right)^{2}}}\)
\(=\frac{1}{\sqrt{5}} \log \left|t+\sqrt{t^{2}-\left(\frac{1}{5}\right)^{2}}\right|+\mathrm{C}\)
\(=\frac{1}{\sqrt{5}} \log \left|x-\frac{1}{5}+\sqrt{x^{2}-\frac{2 x}{5}}\right|+\mathrm{C}\)

(i) We have
∫sin³ x cos²x dx = ∫sin² x cos²x (sin x) dx
= ∫(1 – cos²x) cos²x (sin x) dx
Put t = cos x so that dt = – sin x dx
Therefore, ∫sin²x cos²x (sin x) dx = − ∫(1 – t² ) t² dt
\(=-\int\left(t^{2}-t^{4}\right) d t=-\left(\frac{t^{3}}{3}-\frac{t^{5}}{5}\right)+\mathrm{C} \)
\(=-\frac{1}{3} \cos ^{3} x+\frac{1}{5} \cos ^{5} x+\mathrm{C} \)
(ii) Put x + a = t. Then dx = dt. Therefore
\(\int \frac{\sin x}{\sin (x+a)} d x =\int \frac{\sin (t-a)}{\sin t} d t \)
\(=\int \frac{\sin t \cos a-\cos t \sin a}{\sin t} d t \)
= cos a ∫dt – sin a ∫cot t dt
= (cos a) t – (sin a) ⎡⎣log sin t + C₁⎤⎦
= (cos a) (x + a) – (sin a) ⎡⎣log sin (x + a) + C₁⎤⎦
= x cos a + a cos a – (sin a) log sin (x + a) – C₁ sin a
\(\text { Hence, } \int \frac{\sin x}{\sin (x+a)} d x=x \cos a-\sin a \log |\sin (x+a)|+\mathrm{C}\)
where, C = – C₁ sin a + a cos a, is another arbitrary constant.
\(\text { (iii) } \int \frac{d x}{1+\tan x} =\int \frac{\cos x d x}{\cos x+\sin x} \)
\(=\frac{1}{2} \int \frac{(\cos x+\sin x+\cos x-\sin x) d x}{\cos x+\sin x} \)
\(=\frac{1}{2} \int d x+\frac{1}{2} \int \frac{\cos x-\sin x}{\cos x+\sin x} d x \)
\(=\frac{x}{2}+\frac{C_{1}}{2}+\frac{1}{2} \int \frac{\cos x-\sin x}{\cos x+\sin x} d x \)
\(\text { Now, consider } \mathrm{I}=\int \frac{\cos x-\sin x}{\cos x+\sin x} d x\)
\(\text { Put } \cos x+\sin x=t \text { so that }(\cos x-\sin x) d x=d t\)
\(\text { Therefore } \quad \mathrm{I}=\int \frac{d t}{t}=\log |t|+\mathrm{C}_{2}=\log |\cos x+\sin x|+\mathrm{C}_{2}\)
\(\text { Putting it in (1), we get }\)
\(\int \frac{d x}{1+\tan x} =\frac{x}{2}+\frac{\mathrm{C}_{1}}{2}+\frac{1}{2} \log |\cos x+\sin x|+\frac{\mathrm{C}_{2}}{2} \)
\(=\frac{x}{2}+\frac{1}{2} \log |\cos x+\sin x|+\frac{\mathrm{C}_{1}}{2}+\frac{\mathrm{C}_{2}}{2} \)
\(=\frac{x}{2}+\frac{1}{2} \log |\cos x+\sin x|+\mathrm{C},\left(\mathrm{C}=\frac{\mathrm{C}_{1}}{2}+\frac{\mathrm{C}_{2}}{2}\right) \)