STANDARD INTEGRALS (Excercise Problems with Solutions)

$\LARGE{\color {purple} {PART- A}}$
$\color {purple} {1\ .}\ \color {red} {Evaluate\ :}\ \int \frac{dx}{9\ +\ x^2}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$W.K.T\ \int \frac{dx}{a^2\ +\ x^2} = \frac{1}{a}\ {tan}^{-1} (\frac{x}{a}) +c$
$\int \frac{dx}{9\ +\ x^2}\ = \int \frac{dx}{3^2\ +\ x^2}$
$= \frac{1}{3}\ {tan}^{-1} (\frac{x}{3}) +c$
$\boxed{\int \frac{dx}{9\ +\ x^2} = \frac{1}{3}\ {tan}^{-1} (\frac{x}{2}) +c}$

$\color {purple} {2\ .}\ \color {red} {Evaluate\ :}\ \int \frac{dx}{3\ +\ 2 x^2}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$\int \frac{dx}{3\ +\ 2 x^2} = \int \frac{dx}{2x^2\ +\ 3} = \frac{1}{2}\ \int \frac{dx}{x^2 + \frac{3}{2}} = \frac{1}{2}\ \int \frac{dx}{x^2 + (\sqrt{\frac{3}{2}})^2}$
$W.K.T\ \int \frac{dx}{x^2 + a^2} = \frac{1}{a}\ {tan}^{-1} (\frac{x}{a}) +c$
$\int \frac{dx}{3\ +\ 2 x^2}\ =\ \frac{1}{2}\ × \sqrt{\frac{2}{3}}\ {tan}^{-1} (\frac{x}{\sqrt{\frac{3}{2}}}) +c$
$= \frac{1}{\sqrt{6}}\ {tan}^{-1} (\sqrt{\frac{2}{3}}\ x) +c$
$\boxed{\int \frac{dx}{3\ +\ 2 x^2}\ = \frac{1}{\sqrt{6}}\ {tan}^{-1} (\sqrt{\frac{2}{3}}\ x) +c}$

$\LARGE{\color {purple} {PART- B}}$
$\color {purple} {3\ .}\ \color {red} {Evaluate\ :}\ \int \frac{dx}{{\sqrt{36 – x^2}}}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$\int \frac{dx}{{\sqrt{36\ -\ x^2}}} = \int \frac{dx}{{\sqrt{6^2\ -\ x^2}}}$
$W.K.T\ \int \frac{dx}{{\sqrt{a^2 – x^2}}} = {sin}^{-1}(\frac{x}{a})\ + c$
$\boxed{\int \frac{dx}{{\sqrt{36\ -\ x^2}}} = {sin}^{-1}(\frac{x}{6})\ + c}$

$\color {purple} {4\ .}\ \color {red} {Evaluate\ :}\ \int \frac{dx}{4x^2\ -\ 49}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$\int \frac{dx}{4x^2\ -\ 49} = \frac{1}{4}\ \int \frac{dx}{x^2\ -\ \frac{49}{4}} = \frac{1}{4}\ \int \frac{dx}{x^2\ -\ (\frac{7}{2})^2}$
$W.\ K.\ T\ \int \frac{dx}{x^2 – a^2} = \frac{1}{2a}\ log\ (\frac{x – a}{x + a}) + c$
$\int \frac{dx}{4x^2\ -\ 49}\ =\ \frac{1}{4}[\frac{1}{2\ ×\ \frac{7}{2}}\ log\ (\frac{x\ -\ \frac{7}{2}}{x\ +\ \frac{7}{2}})] +c$
$=\ \frac{1}{28}\ log\ (\frac{2x\ -\ 7}{2x\ +\ 7}) + c$
$\boxed{\int \frac{dx}{4x^2\ -\ 49} = \frac{1}{28}\ log\ (\frac{2x\ -\ 7}{2x\ +\ 7}) + c}$

$\LARGE{\color {purple} {PART- C}}$
$\color {purple} {5\ .}\ \color {red} {Evaluate\ :}\ \hspace{2cm}\ (i)\ \int\ \frac{dx}{16\ +\ x^2}\ \hspace{2cm}\ (ii)\ \int\frac{dx}{{\sqrt{4\ -\ (x\ +\ 1)^2}}}\ \hspace{10cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$(i)\ \int\ \frac{dx}{16\ +\ x^2}\ dx\ \hspace{10cm}$
$W.K.T\ \int \frac{dx}{a^2\ +\ x^2} = \frac{1}{a}\ {tan}^{-1} (\frac{x}{a}) +c$
$\int \frac{dx}{16\ +\ x^2}\ = \int \frac{dx}{4^2\ +\ x^2}$
$= \frac{1}{4}\ {tan}^{-1} (\frac{x}{4}) +c$
$\boxed{\int \frac{dx}{16\ +\ x^2} = \frac{1}{4}\ {tan}^{-1} (\frac{x}{4}) +c}$

$(ii)\ \int\ \frac{dx}{{\sqrt{4\ -\ (x\ +\ 1)^2}}}\ \hspace{10cm}$
$Put\ u\ =\ x\ +\ 1$
$\frac{du}{dx}= \ 1$
$dx = \ du$
$\int \frac{dx}{{\sqrt{4\ -\ (x\ +\ 1)^2}}} = \ \int \frac{du}{{\sqrt{2^2\ -\ u^2}}}$
$W.K.T\ \int \frac{dx}{{\sqrt{a^2 – x^2}}} = {sin}^{-1}(\frac{x}{a})\ + c$
$\int \frac{dx}{{\sqrt{4\ -\ (x\ +\ 1)^2}}}\ =\ {sin}^{-1}(\frac{u}{2})$
$=\ {sin}^{-1}(\frac{x\ +\ 1}{2})\ + c$
$\boxed{\int \frac{dx}{{\sqrt{4\ -\ (x\ +\ 1)^2}}}\ = {sin}^{-1}(\frac{x\ +\ 1}{2})\ + c}$

$\color {purple} {6\ .}\ \color {red} {Evaluate\ :}\ \hspace{2cm}\ (i)\ \int\ \frac{dx}{(7x + 1)^2 + 25}\ \hspace{2cm}\ (ii)\ \int \frac{dx}{{\sqrt{4\ -\ 81x^2}}}\ \hspace{10cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$(i)\ \int\ \frac{dx}{(7x + 1)^2 + 25}\ \hspace{10cm}$
$Put\ u\ =\ 7x\ +\ 1$
$\frac{du}{dx}= \ 7$
$dx = \frac{1}{7}\ du$
$\int \frac{dx}{(7x + 1)^2 + 25} = \frac{1}{7}\ \int \frac{du}{u^2 + 5^2}$
$W.K.T\ \int \frac{dx}{x^2 + a^2} = \frac{1}{a}\ {tan}^{-1} (\frac{x}{a}) +c$
$\frac{1}{7}\ \int \frac{du}{u^2 + 5^2} = \frac{1}{3}\ × \frac{1}{7}\ {tan}^{-1} (\frac{u}{5}) +c$
$= \frac{1}{35}\ {tan}^{-1} (\frac{7x + 1}{5}) +c$
$\boxed{\int \frac{dx}{(7x + 1)^2\ +\ 25} = \frac{1}{35}\ {tan}^{-1} (\frac{7x + 1}{5}) +c}$

$(ii)\ \int\ \frac{dx}{{\sqrt{4\ -\ 81x^2}}}\ \hspace{10cm}$
$\int \frac{dx}{{\sqrt{4\ -\ 81x^2}}} = \frac{1}{9}\ \int \frac{dx}{{\sqrt{\frac{4}{81} – x^2}}}$
$= \frac{1}{9}\ \int \frac{dx}{{\sqrt{(\frac{2}{9})^2 – x^2}}}$
$W.K.T\ \int \frac{dx}{{\sqrt{a^2 – x^2}}} = {sin}^{-1}(\frac{x}{a})\ + c$
$\frac{1}{9}\ \int \frac{dx}{{\sqrt{(\frac{2}{9})^2 – x^2}}}\ =\ \frac{1}{9}\ {sin}^{-1}(\frac{x}{\frac{2}{9}})\ + c$
$= \frac{1}{9}\ {sin}^{-1}(\frac{9x}{2})\ + c$
$\boxed{\int\ \frac{dx}{{\sqrt{4\ -\ 81x^2}}}\ =\ \frac{1}{9}\ {sin}^{-1}(\frac{9x}{2})\ + c}$

$\color {purple} {7\ .}\ \color {red} {Evaluate\ :}\ \hspace{2cm}\ (i)\ \int\ \frac{dx}{64\ -\ x^2}\ \hspace{2cm}\ (ii)\ \int\frac{dx}{{\sqrt{36\ -\ (5x\ +\ 1)^2}}}\ \hspace{10cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$(i)\ \int\ \frac{dx}{64\ -\ x^2}\ \hspace{10cm}$
$\int \frac{dx}{64\ -\ x^2}\ = \ \int \frac{dx}{8^2\ -\ x^2}$
$W.K.T\ \int \frac{dx}{a^2 – x^2} = \frac{1}{2a}\ log\ (\frac{a + x}{a – x}) + c$
$\int \frac{dx}{64\ -\ x^2}\ = \ \frac{1}{2 × 8 }\ log\ (\frac{8\ +\ x}{8\ -\ x})\ +c$
$=\ \frac{1}{16}\ log\ (\frac{8\ +\ x}{8\ -\ x})\ + c$
$\boxed{\int \frac{dx}{64\ -\ x^2}\ = \frac{1}{16}\ log\ (\frac{8\ +\ x}{8\ -\ x})\ + c}$

$(ii)\ \int\ \frac{dx}{{\sqrt{36\ -\ (5x\ +\ 1)^2}}}\ \hspace{10cm}$
$Put\ u\ =\ 5x\ +\ 1$
$\frac{du}{dx}= \ 5$
$dx =\ \frac{1}{5} \ du$
$\int \frac{dx}{{\sqrt{36\ -\ (5x\ +\ 1)^2}}} = \ \frac{1}{5}\int \frac{du}{{\sqrt{6^2\ -\ u^2}}}$
$W.K.T\ \int \frac{dx}{{\sqrt{a^2 – x^2}}} = {sin}^{-1}(\frac{x}{a})\ + c$
$\int \frac{dx}{{\sqrt{36\ -\ (5x\ +\ 1)^2}}}\ =\ \frac{1}{5}\ {sin}^{-1}(\frac{u}{6})$
$=\ \frac{1}{5}\ {sin}^{-1}(\frac{5x\ +\ 1}{6})\ + c$
$\boxed{\int \frac{dx}{{\sqrt{36\ -\ (5x\ +\ 1)^2}}}\ = \frac{1}{5}\ {sin}^{-1}(\frac{5x\ +\ 1}{6})\ + c}$

$\color {purple} {8\ .}\ \color {red} {Evaluate\ :}\ \int\frac{dx}{(2x + 3)^2\ +\ 49}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$\int\ \frac{dx}{(2x + 3)^2\ +\ 49}\ \hspace{10cm}$
$Put\ u\ =\ 2x\ +\ 3$
$\frac{du}{dx}= \ 2$
$dx = \frac{1}{2}\ du$
$\int \frac{dx}{(2x + 3)^2\ +\ 49} = \frac{1}{2}\ \int \frac{du}{u^2\ +\ 7^2}$
$W.K.T\ \int \frac{dx}{x^2 + a^2} = \frac{1}{a}\ {tan}^{-1} (\frac{x}{a}) +c$
$\int \frac{dx}{(2x + 3)^2\ +\ 49} = \frac{1}{2}\ × \frac{1}{7}\ {tan}^{-1} (\frac{u}{7}) +c$
$= \frac{1}{14}\ {tan}^{-1} (\frac{2x\ +\ 3}{7}) +c$
$\boxed{\int \frac{dx}{(2x + 3)^2\ +\ 49} = \frac{1}{14}\ {tan}^{-1} (\frac{2x\ +\ 3}{7}) +c}$

$\color {purple} {9\ .}\ \color {red} {Evaluate\ :}\ \hspace{2cm}\ (i)\ \int\ \frac{dx}{25\ -\ 9x^2}\ \hspace{2cm}\ (ii)\ \int\frac{dx}{(2x + 3)^2\ +\ 9}\ \hspace{10cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$(i)\ \int\ \frac{dx}{25\ -\ 9x^2}\ \hspace{10cm}$
$\int \frac{dx}{25\ -\ 9x^2} = \frac{1}{9}\ \int \frac{dx}{\frac{25}{9}\ -\ x^2} = \frac{1}{9}\ \int \frac{dx}{(\frac{5}{3})^2\ – x^2}$
$W.K.T\ \int \frac{dx}{a^2 – x^2} = \frac{1}{2a}\ log\ (\frac{a + x}{a – x}) + c$
$\int \frac{dx}{25\ -\ 9x^2}\ = \ \frac{1}{9}[\frac{1}{2 × \frac{5}{3}}\ log\ (\frac{\frac{5}{3}\ +\ x}{\frac{5}{3}\ -\ x})\ +c]$
$=\ \frac{1}{30}\ log\ (\frac{5\ +\ 3x}{5\ -\ 3x})\ + c$
$\boxed{\int \frac{dx}{25\ -\ 9x^2}\ = \frac{1}{30}\ log\ (\frac{5\ +\ 3x}{5\ -\ 3x})\ + c}$

$(ii)\ \int\ \frac{dx}{(2x + 3)^2\ +\ 9}\ \hspace{10cm}$
$Put\ u\ =\ 2x\ +\ 3$
$\frac{du}{dx}= \ 2$
$dx = \frac{1}{2}\ du$
$\int \frac{dx}{(2x + 3)^2\ +\ 9} = \frac{1}{2}\ \int \frac{du}{u^2 + 3^2}$
$W.K.T\ \int \frac{dx}{x^2 + a^2} = \frac{1}{a}\ {tan}^{-1} (\frac{x}{a}) +c$
$\int \frac{dx}{(2x + 3)^2\ +\ 9} = \frac{1}{2}\ × \frac{1}{3}\ {tan}^{-1} (\frac{u}{3}) +c$
$= \frac{1}{6}\ {tan}^{-1} (\frac{2x\ +\ 3}{3}) +c$
$\boxed{\int \frac{dx}{(2x + 3)^2\ +\ 9} = \frac{1}{6}\ {tan}^{-1} (\frac{2x\ +\ 3}{3}) +c}$

$\color {purple} {10\ .}\ \color {red} {Evaluate\ :}\ \int\ \frac{dx}{3\ -\ 2x\ -\ x^2}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$3\ -\ 2x\ -\ x^2\ =\ 3\ -\ [(x\ +\ 1)^2\ -\ 1]$
$=\ 4\ – (x\ +\ 1)^2$
$\int \frac{dx}{3\ -\ 2x\ -\ x^2}\ =\ \int \frac{dx}{4\ – (x\ +\ 1)^2}$
$Put\ u\ =\ x\ +\ 1$
$\frac{du}{dx}= \ 1$
$dx = \ du$
$\int \frac{dx}{4\ – (x\ +\ 1)^2}\ =\ \int \frac{du}{2^2 – u^2}$
$W.K.T\ \int \frac{dx}{a^2 – x^2} = \frac{1}{2a}\ log\ (\frac{a + x}{a – x}) + c$
$\int \frac{dx}{3\ -\ 2x\ -\ x^2} = \ \frac{1}{2 × 2}\ log\ (\frac{2 + u}{2 – u})\ +c$
$=\frac{1}{4}\ log\ (\frac{2 + (x\ +\ 1)}{2\ -\ (x\ +\ 1)}) +c$
$=\frac{1}{4}\ log\ (\frac{2\ +\ x\ +\ 1}{2\ -\ x\ -\ 1})] +c$
$=\frac{1}{4}\ log\ (\frac{3\ +\ x }{1\ -\ x })\ +c$
$\boxed{\int \frac{dx}{3\ -\ 2x\ -\ x^2} = \frac{1}{4}\ log\ (\frac{3\ +\ x }{1\ -\ x })\ + c}$