Integral Calculus I Exercise 2.7 - 12th Text Book Solutions

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Important Formulas in Integral Calculus I

Text Book Solutions for Integral Calculus I Exercise 2.1

Text Book Solutions for Integral Calculus I Exercise 2.2

Text Book Solutions for Integral Calculus I Exercise 2.3

Text Book Solutions for Integral Calculus I Exercise 2.4

Text Book Solutions for Integral Calculus I Exercise 2.5

Text Book Solutions for Integral Calculus I Exercise 2.6

Text Book Solutions for Integral Calculus I Exercise 2.8

Text Book Solutions for Integral Calculus I Exercise 2.9

Text Book Solutions for Integral Calculus I Exercise 2.10

Text Book Solutions for Integral Calculus I Exercise 2.11

Text Book Solutions for Integral Calculus I Exercise 2.12 (MCQ)

Integral Calculus I Exercise 2.7 Text Book Solutions

Integrate the following with respect to x

$1. \: \dfrac{1}{9-16x^{2}}$

$Let \: I =\int \dfrac{1}{9-16x^{2}}\: dx\\$
$=\int \dfrac{1}{(3)^{2}-(4x)^{2}} \: dx\\$
$Using \: formula \: \int \dfrac{dx}{a^{2}-x^{2}}=\dfrac{1}{2a}\: log \left | \dfrac{a+x}{a-x} \right |+c\\$
$=\dfrac{1}{2(3)}\:\frac{ log \left | \dfrac{3+4x}{3-4x} \right |}{4}+c\\$
$=\dfrac{1}{24}\: log \left | \dfrac{3+4x}{3-4x} \right |+c$

$2. \: \dfrac{1}{9-8x-x^{2}}$

$Let \:I = \int \dfrac{1}{9-8x-x^{2}}\: dx \\$
$9-8x-x^{2}=-x^{2}-8x+9\\$
$=-(x^{2}+8x-9)\\$
$=-(x^{2}+8x+4^{2}-4^{2}-9) \: (Using \: completion \: of \: squares\: method)\\$
$=-(x^{2}+8x+4^{2}-16-9)\\$
$=-(x+4)^{2}-25\\$
$=25-(x+4)^{2}\\$
$=5^{2}-(x+4)^{2}\\$
$Using \: formula \: \int \dfrac{dx}{a^{2}-x^{2}}=\dfrac{1}{2a}\: log \left | \dfrac{a+x}{a-x} \right |+c\\$
$I =\int \dfrac{dx}{5^{2}-(x+4)^{2}}\\$
$=\dfrac{1}{2(5)}\: log \left | \dfrac{5+x+4}{5-x-4} \right |+c\\$
$=\dfrac{1}{10}\: log \left | \dfrac{9+x}{1-x} \right |+c$

$3. \: \dfrac{1}{2x^{2}-9}$

$Let \: I =\int \dfrac{1}{2x^{2}-9}\: dx\\$
$=\int \dfrac{dx}{(\sqrt{2x})^{2}-(3)^{2}}\\$
$Using \: formula \: \int \dfrac{dx}{x^{2}-a^{2}}=\dfrac{1}{2a}\: log \left | \dfrac{x-a}{x+a} \right |+c\\$
$=\dfrac{1}{2(3)}\: \dfrac{log \left | \dfrac{(\sqrt{2x})-3}{(\sqrt{2x})+3} \right |}{\sqrt{2}}+c\\$
$=\dfrac{1}{6(\sqrt{2})}\: log \left | \dfrac{(\sqrt{2x})-3}{(\sqrt{2x})+3} \right |+c$

$4. \: \dfrac{1}{x^{2}-x-2}$

$Let \: I =\int \dfrac{1}{x^{2}-x-2}\: dx\\$
$Using \: completion \: of \: squares\: method\\$
$x^{2}-x-2=x^{2}-x+\left (\frac{1}{2} \right )^{2} - \left(\frac{1}{2} \right )^{2}-2 \\$
$=\left (x-\frac{1}{2} \right )^{2}-\frac{1}{4}-2\\$
$=\left (x-\frac{1}{2} \right )^{2}-\frac{9}{4}\\$
$=\left (x-\frac{1}{2} \right )^{2}-\left (\frac{3}{2} \right )^{2}$
$I=\int \dfrac{dx}{\left (x-\frac{1}{2} \right )^{2}-\left (\frac{3}{2} \right )^{2}}\\$
$Using \: formula \: \int \dfrac{dx}{x^{2}-a^{2}}=\dfrac{1}{2a}\: log \left | \dfrac{x-a}{x+a} \right |+c\\$
$=\dfrac{1}{2\left ( \dfrac{3}{2} \right )}log \left | \dfrac{x-\dfrac{1}{2}-\dfrac{3}{2}}{x-\dfrac{1}{2}+\dfrac{3}{2}} \right |+c\\$
$=\dfrac{1}{3}\: log \left | \dfrac{x-2}{x+1} \right |+c$

$5. \: \dfrac{1}{x^{2}+3x+2}$

$Let \: I = \int \dfrac{dx}{x^{2}+3x+2} \\$
$Using\: completion\: of\: squares\: method \\$
$x^{2}+3x+2=x^{2}+3x+\left ( \frac{3}{2} \right )^{2}-\left ( \frac{3}{2} \right )^{2}+2 \\$
$=\left ( x+\dfrac{3}{2} \right )^{2}-\dfrac{9}{4}+2\\$
$=\left ( x+\dfrac{3}{2} \right )^{2}-\dfrac{1}{4}\\$
$=\left ( x+\dfrac{3}{2} \right )^{2}-\left (\dfrac{1}{2} \right )^{2}\\$
$I =\int \dfrac{dx}{\left ( x+\dfrac{3}{2} \right )^{2}-\left (\dfrac{1}{2} \right )^{2}}\\$
$Using \: formula \: \int \dfrac{dx}{x^{2}-a^{2}}=\dfrac{1}{2a}\: log \left | \dfrac{x-a}{x+a} \right |+c\\$
$=\dfrac{1}{2\left (\dfrac{1}{2} \right )}\: log \left | \dfrac{x+\frac{3}{2}-\frac{1}{2}}{x-\dfrac{3}{2}+\frac{1}{2}} \right |+c\\$
$=log\left | \frac{x+1}{x+2} \right |+c$

$6. \: \dfrac{1}{2x^{2}+6x-8}$

$Let\: I = \int \dfrac{dx}{2x^{2}+6x-8} \\$
$2x^{2}+6x-8 = 2\left ( x^{2} +3x-4\right )\\$
$=2\left ( x^{2} +3x+\left (\dfrac{3}{2} \right )^{2}- \left (\dfrac{3}{2} \right )^{2}-4\right )\\$
$=2\left [ \left (x+\dfrac{3}{2} \right )^{2} -\dfrac{9}{4}-4\right ]\\$
$=2\left [ \left (x+\dfrac{3}{2} \right )^{2} -\dfrac{25}{4}\right ]\\$
$=2\left [ \left (x+\dfrac{3}{2} \right )^{2} -\left (\dfrac{5}{2} \right )^{2}\right ]\\$
$I=\int \dfrac{dx}{2\left [ \left (x+\dfrac{3}{2} \right )^{2} -\left (\dfrac{5}{2} \right )^{2}\right ]}\\$
$=\dfrac{1}{2}\int \dfrac{dx}{\left (x+\dfrac{3}{2} \right )^{2}-\left (\dfrac{5}{2} \right )^{2}}\\$
$=\dfrac{1}{2}\left ( \dfrac{1}{2\times \dfrac{5}{2}} \right ) \: log \left | \dfrac{x+\dfrac{3}{2}-\dfrac{5}{2}}{{x+\dfrac{3}{2}+\dfrac{5}{2}}} \right |+c\\$
$=\dfrac{1}{10} \: log \left | \dfrac{x-1}{x+4} \right |+c$

$7. \: \dfrac{e^{x}}{e^{2x}-9}$

$I = \int \dfrac{e^{x}}{e^{2x}-9}\: dx\\$
$= \int \dfrac{e^{x}dx}{(e^{x})^{2}-(3)^{2}}\\$
$Put \: t= e^{x}\\$
$dt = e^{x} \: dx\\$
$I = \int \dfrac{dt}{t^{2}-3^{2}}\\$
$Using \: formula \: \int \dfrac{dx}{x^{2}-a^{2}}=\dfrac{1}{2a}\: log \left | \dfrac{x-a}{x+a} \right |+c\\$
$=\dfrac{1}{2\times 3} \: log\left | \dfrac{t-3}{t+3} \right |+c\\$
$=\dfrac{1}{6} \: log\left | \dfrac{e^{x}-3}{e^{x}+3} \right |+c$

$8. \: \dfrac{1}{\sqrt{9x^{2}-7}}$

$I = \dfrac{dx}{\sqrt{9x^{2}-7}}\\$
$= \dfrac{dx}{\sqrt{(3x)^{2}-(\sqrt{7})^{2}}}\\$
$Using \: formula \: \int \dfrac{dx}{x^{2}-a^{2}} = log\left | x+ \sqrt{x^{2}-a^{2}}\right |+c\\$
$I =\dfrac{log\left | 3x+ \sqrt{9x^{2}-7}\right |}{3}+c$

$9. \: \dfrac{1}{\sqrt{x^{2}+6x+13}}$

$I =\int \dfrac{dx}{\sqrt{x^{2}+6x+13}}\\$
$x^{2}+6x+13 = x^{2}+6x+3^{2}-3^{2}+13\\$
$=(x+3)^{2}-9+13\\$
$=(x+3)^{2}+4\\$
$=(x+3)^{2}+2^{2}\\$
$I=\int \dfrac{dx}{\sqrt{(x+3)^{2}+2^{2}}}\\$
$=log \left |(x+3)+{\sqrt{(x+3)^{2}+2^{2}}} \right |+c\\$
$=log \left |(x+3)+{\sqrt{x^{2}+6x+13}} \right |+c$

$10. \: \dfrac{1}{\sqrt{x^{2}-3x+2}}$

$I =\int \dfrac{dx}{\sqrt{x^{2}-3x+2}}\\$
$x^{2}-3x+2=x^{2}-3x+\left ( \dfrac{3}{2} \right )^{2}-\left ( \dfrac{3}{2} \right )^{2}+2\\$
$=\left ( x-\dfrac{3}{2} \right )^{2}-\dfrac{9}{4}+2\\$
$=\left ( x-\dfrac{3}{2} \right )^{2}-\dfrac{1}{4}\\$
$=\left ( x-\dfrac{3}{2} \right )^{2}-\left (\dfrac{1}{2} \right )^{2}\\$
$I=\int \dfrac{dx}{\left ( x-\dfrac{3}{2} \right )^{2}-\left (\dfrac{1}{2} \right )^{2}}\\$
$=log \left |\left (x-\dfrac{3}{2} \right )+\sqrt{{\left ( x-\dfrac{3}{2} \right )^{2}-\left (\dfrac{1}{2} \right )^{2}}} \right |+c\\$
$=log \left |\left (x-\dfrac{3}{2} \right )+\sqrt{x^{2}-3x+2} \right |+c$

$11. \: \dfrac{x^{3}}{\sqrt{{x^{8}-1}}}$

$I = \int \dfrac{x^{3}}{\sqrt{{x^{8}-1}}}\: dx\\$
$=\int \dfrac{x^{3}}{\sqrt{(x^{4})^{2}-1^{2}}}\: dx\\$
$Put \: t=x^{4}\\$
$dt=4x^{3}dx\\$
$\frac{dt}{4}=x^{3}dx\\$
$I= \int \dfrac{\dfrac{dt}{4}}{\sqrt{t^{2}-1}}\\$
$=\dfrac{1}{4}\int \dfrac{dt}{\sqrt{t^{2}-1}}\\$
$Using \: formula \: \int \dfrac{dx}{x^{2}-a^{2}} = log\left | x+ \sqrt{x^{2}-a^{2}}\right |+c\\$
$=\dfrac{1}{4} log \left |t+\sqrt{t^{2}-1} \right |+c\\$
$=\dfrac{1}{4} log \left |x^{4}+\sqrt{x^{8}-1} \right |+c$

$12. \: \sqrt{1+x+x^{2}}$

$I =\int \sqrt{1+x+x^{2}} \: dx\\$
$1+x+x^{2}=x^{2}+x+1\\$
$=x^{2}+x+\left ( \dfrac{1}{2} \right )^{2}-\left ( \dfrac{1}{2} \right )^{2}+1\\$
$=\left (x+\dfrac{1}{2} \right )^{2}-\dfrac{1}{4}+1\\$
$=\left (x+\dfrac{1}{2} \right )^{2}+\dfrac{3}{4}\\$
$=\left (x+\dfrac{1}{2} \right )^{2}+\left (\dfrac{\sqrt{3}}{2} \right )^{2}\\$
$=\left (x+\dfrac{1}{2} \right )^{2}+\left (\dfrac{\sqrt{3}}{2} \right )^{2}\\$
$I=\int \sqrt{\left (x+\dfrac{1}{2} \right )^{2}+\left (\dfrac{\sqrt{3}}{2} \right )^{2}}\: dx\\$
$Using \: formula\: \int \sqrt{x^{2}+a^{2}}\: dx =\dfrac{x}{2}\sqrt{x^{2}+a^{2}}+\dfrac{a^{2}}{2}log \left | x+\sqrt{x^{2}+a^{2}} \right |+c\\$
$\dfrac{\left (x+\dfrac{1}{2} \right )}{2}\sqrt{\left (x+\dfrac{1}{2} \right )^{2}+\left (\dfrac{\sqrt{3}}{2} \right )^{2}}+\dfrac{\left (\dfrac{\sqrt{3}}{2} \right )^{2}}{2} log \left |\left ( x+\dfrac{1}{2} \right )+\sqrt{\left (x+\dfrac{1}{2} \right )^{2}+\left (\dfrac{\sqrt{3}}{2} \right )^{2}} \right |+c\\$
$I=\dfrac{x+\dfrac{1}{2}}{2}\sqrt{1+x+x^{2}}+\dfrac{3}{8}\:log \left |\left ( x+\dfrac{1}{2} \right )+\sqrt{ 1+x+x^{2}} \right |+c\\$
$=\dfrac{2x+1}{4}\sqrt{1+x+x^{2}}+\dfrac{3}{8}\:log \left |\left ( x+\dfrac{1}{2} \right )+\sqrt{ 1+x+x^{2}} \right |+c$

$13. \: \sqrt{x^{2}-2}$

$I = \int \sqrt{x^{2}-2}\: dx\\$
$=\int \sqrt{x^{2}-\sqrt{2}^{2}}\: dx\\$
$=\dfrac{x}{2}\sqrt{x^{2}-\sqrt{2}^{2}} -\dfrac{\sqrt{2}^{2}}{2}log\left | x+ \sqrt{x^{2}-\sqrt{2}^{2}} \right |+c\\$
$=\dfrac{x}{2}\sqrt{x^{2}-2 } - log\left | x+ \sqrt{x^{2}-2} \right |+c$

$14. \sqrt{4x^{2}-5}$

$I = \int \sqrt{4x^{2}-5}\: dx\\$
$=\int \sqrt{(2x)^{2}-\sqrt{5}^{2}}\: dx\\$
$=\dfrac{\dfrac{2x}{2}\sqrt{(2x)^{2}-\sqrt{5}^{2}}-\dfrac{\sqrt{5}^{2}}{2}log\left |\: 2x+\sqrt{(2x)^{2}-\sqrt{5}^{2}} \right |+c}{2}$
$=\dfrac{1}{2} \left [x\sqrt{4x^{2}-5}-\dfrac{5}{2}log \left |2x+\sqrt{4x^{2}-5} \right | \right ]+c\\$
$=\dfrac{1}{2} \left [2x\sqrt{4x^{2}-5}-5\: log \left |2x+\sqrt{4x^{2}-5} \right | \right ]+c$

$15. \sqrt{2x^{2}+4x+1}$

$I = \int \sqrt{2x^{2}+4x+1} \: dx\\$
$2x^{2}+4x+1 = 2\left ( x^{2}+2x+\dfrac{1}{2} \right )\\$
$Using\: completion \: of \: squares,\\$
$= 2\left ( x^{2}+2x+1^{2}-1^{2}+\dfrac{1}{2} \right )\\$
$=2 \left [\left ( x+1 \right )^{2}-1+\dfrac{1}{2}\right ]\\$
$=2 \left [\left ( x+1 \right )^{2}-\dfrac{1}{2}\right ]\\$
$=2 \left [\left ( x+1 \right )^{2}-\left (\dfrac{1}{\sqrt{2}} \right )^{2}\right ]\\$
$I = \int \sqrt{2 \left [\left ( x+1 \right )^{2}-\left (\dfrac{1}{\sqrt{2}} \right )^{2}\right ]}\: dx\\$
$=\sqrt{2\int }\sqrt{ \left [\left ( x+1 \right )^{2}-\left (\dfrac{1}{\sqrt{2}} \right )^{2}\right ]}\: dx\\$
$=\sqrt{2\int }\sqrt{ \left [\left ( x+1 \right )^{2}-\left (\dfrac{1}{\sqrt{2}} \right )^{2}\right ]}\: dx\\$
$\sqrt{2}\left (\dfrac{x+1}{2} \right )\sqrt{ \left [\left ( x+1 \right )^{2}-\left (\dfrac{1}{\sqrt{2}} \right )^{2}\right ]}-\dfrac{\left (\frac{1}{\sqrt{2}} \right )^{2}}{2} log \left |\left ( x+1 \right ) + \sqrt{\left (x+1 \right )^{2}- \left (\dfrac{1}{\sqrt{2}} \right )^{2}} \right |+c\\$
$\sqrt{2} \left (\dfrac{x+1}{2} \right ) \left (x^{2}+2x+\dfrac{1}{2} \right )-\dfrac{1}{4}log \left |\left (x+1 \right )+\sqrt{\left (x^{2}+2x+\dfrac{1}{2} \right )} \right |+c$

$16. \dfrac{1}{x+\sqrt{x^{2}-1}}$

$I =\int \dfrac{dx}{x+\sqrt{x^{2}-1}}\\$
$= \dfrac{1}{x+\sqrt{x^{2}-1}} =\dfrac{1}{x+\sqrt{x^{2}-1}} \times \dfrac{x-\sqrt{x^{2}-1}}{x-\sqrt{x^{2}-1}}\\$
$=\dfrac{x-\sqrt{x^{2}-1}}{ x^{2} -\left (x^{2}-1 \right )}\\$
$=\dfrac{x-\sqrt{x^{2}-1}}{ x^{2} -x^{2}+1}\\$
$=x-\sqrt{x^{2}-1}\\$
$I = \int\left ( x-\sqrt{x^{2}-1} \right )\: dx\\$
$=\int\left ( x-\sqrt{x^{2}-1^{2}} \right )\: dx\\$
$=\dfrac{x^{2}}{2}-\left [ \dfrac{x}{2} \right ]\sqrt{x^{2}-1^{2}}-\dfrac{1^{2}}{2}log \left |x+\sqrt{x^{2}-1} \right |+c\\$
$=\dfrac{x^{2}}{2}-\left [ \dfrac{x}{2} \right ]\sqrt{x^{2}-1}-\dfrac{1}{2}log \left |x+\sqrt{x^{2}-1} \right |+c$