(i) Find the following integral: \( \int \dfrac{x+2}{2x^2+6x+5}.dx\)
(ii) Find the following integral: \( \int \dfrac{x+3}{\sqrt{5-4x-x^2}}.dx\)
(ii) Find the following integral: \( \int \dfrac{x+3}{\sqrt{5-4x-x^2}}.dx\)
(i)To find: \( \int \dfrac{x+2}{2x^2+6x+5}.dx\)
We can write numerator as
\( \Rightarrow x+2 = A \dfrac{d}{dx} ( 2x^2 +6x+5)+B ...(1)\)
\( \Rightarrow x+2 = A(4x+6)+B\)
\( \Rightarrow x+2 = 4Ax+(6A+B)\)
On comparing coefficient of \(x\) , we get
\(1= 4A \Rightarrow A = \dfrac{1}{4} \)
Comparing constant term
\( \Rightarrow 2 = 6A + B \Rightarrow \dfrac{6}{4} + B =2 \)
\( \Rightarrow B = 2- \dfrac{3}{2} = \dfrac{1}{2} \)
So,
\( A = \dfrac{1}{4} \) and \( B = \dfrac{1}{2} \)
Putting the value of \(A\) and \(B\) in \((1)\) , we get
\( \Rightarrow x+2 = \dfrac{1}{4}(4x+6) + \dfrac{1}{2} \)
Now, \( \int \dfrac{x+2}{2x^2 +6x+5}.dx = \int \dfrac{\dfrac{1}{4} (4x+6) + \dfrac{1}{2}}{2x^2 +6x+5}.dx = \dfrac{1}{4} \int \dfrac{4x+6}{2x^2 +6x+5\over{I_1}}.dx +\dfrac{1}{2} \int \dfrac{1}{2x^2 +6x+5\over{I_2}}.dx \) ....(2)
Solving \(I_1\)
\( \Rightarrow I_1= \dfrac{1}{4} \int \dfrac{4x+6}{2x^2 +6x+5}.dx \)
Taking \( t= 2x^2 +6x+5 \)
Differentiating both sides w.r.t.\(x\)
\( \Rightarrow \dfrac{dt}{dx}= (4x+6) \Rightarrow dt = (4x+6)dx \)
Thus,
\( I_1 = \int \dfrac{4x+6}{2x^2 +6x+5}.dx \)
\( \Rightarrow I_1 = \dfrac{1}{4} \int \dfrac{1}{t}.dt \)
\( \Rightarrow I_1 = \dfrac{1}{4} \text{log} |t| + C_1\)
\( \Rightarrow I_1 = \dfrac{1}{4} \text{log} |2x^2 +6x+5 | + C_1\)
Solving \( I_2\)
\( I_2 = \dfrac{1}{2} \int \dfrac{1}{2x^2 +6x+5}.dx \)
\( \Rightarrow I_2 = \dfrac{1}{2} \int \dfrac{1}{2\left [ x^2 + \dfrac{6x}{2} + \dfrac{5}{2}\right ]}.dx \)
\( \Rightarrow I_2 = \dfrac{1}{4} \int \dfrac{1}{x^2 +3x + \dfrac{5}{2} }.dx \)
\( \Rightarrow I_2 = \dfrac{1}{4} \int \dfrac{1}{x^2 +3x +\left ( \dfrac{3}{2} \right )^2 - \left ( \dfrac{3}{2} \right )^2+ \dfrac{5}{2} }.dx \)
\( \Rightarrow I_2 = \dfrac{1}{4} \int \dfrac{1}{\left ( x+ \dfrac{3}{2} \right )^2 -\dfrac{9}{4} + \dfrac{5}{2} }.dx \)
\( \Rightarrow I_2 = \dfrac{1}{4} \int \dfrac{1}{\left ( x+ \dfrac{3}{2} \right )^2 + \dfrac{1}{4} }.dx \)
\( \Rightarrow I_2 = \dfrac{1}{4} \int \dfrac{1}{\left ( x+ \dfrac{3}{2} \right )^2 + \left ( \dfrac{1}{2} \right )^2 }.dx \)
\( \Rightarrow I_2 = \dfrac{1}{4}\left [ \dfrac{1}{\dfrac{1}{2}} \tan^{-1} \left ( \dfrac{x+ \dfrac{3}{2}}{\dfrac{1}{2}} \right ) \right ] + C_2 \)
\( \left [ \because \int \dfrac{dx}{x^2+a^2}= \dfrac{1}{a} \tan^{-1}\dfrac{x}{a} +C \right ] \)
\( \Rightarrow I_2 = \dfrac{1}{4} [ 2~\tan^{-1} (2x+3)]+C_2\)
\( \Rightarrow I_2 = \dfrac{1}{2} \tan^{-1} (2x+3)+C_2\)
Now, putting the value of \( I_1\) and \(I_2\) in equation \((2)\)
\( \therefore \int \dfrac{x+2}{2x^2 +6x+5 }.dx\)
\( = \dfrac{1}{4} \text{log} |2x^2 +6x+5 | + C_1 +\dfrac{1}{2} \tan^{-1} (2x+3)+C_2 \)
\( = \dfrac{1}{4} \text{log} |2x^2 +6x+5 | +\dfrac{1}{2} \tan^{-1} (2x+3)+ C \)
(ii)\( \int \dfrac{x+3}{\sqrt{5-4x-x^2}}.dx\)
We can write numerator as
\( \Rightarrow x+3 = A \dfrac{d}{dx} ( -x^2 -4x+5)+B ...(1)\)
\( \Rightarrow x+3 = A(-2x-4)+B\)
\( \Rightarrow x+3 = -2Ax-4A+B\)
On comparing coefficient of \(x\) , we get
\( \Rightarrow 1= -2A \Rightarrow A = \dfrac{-1}{2} \)
Comparing constant term
\( \Rightarrow 3 = -4A + B \)
\( \Rightarrow B = -4 \left ( \dfrac{-1}{2} \right )+B\)
\( \Rightarrow B = 3-2=1\)
So,
\( A = \dfrac{-1}{2} \) and \( B = 1 \)
Putting the value of \(A\) and \(B\) in \((1)\) , we get
\( \Rightarrow x+3 = \dfrac{-1}{2}(-2x-4) +1 \)
Now, \( \int \dfrac{x+3}{\sqrt{5-4x-x^2}}.dx = \int \dfrac{\dfrac{-1}{2} (-2x-4) + 1}{\sqrt{5-4x-x^2} }.dx = \dfrac{-1}{4} \int \dfrac{-2x-6}{\sqrt{5-4x-x^2} \over{I_1}}.dx + \int \dfrac{1}{\sqrt{5-4x-x^2} \over{I_2}}.dx \) ....(2)
Solving \(I_1\)
\( \Rightarrow I_1= \dfrac{-1}{2} \int \dfrac{-2x-4}{ \sqrt{-x^2-4x+5}}.dx \)
Let \( t= -x^2 -4x+5\)
Differentiating both sides w.r.t.\(x\)
\( \Rightarrow \dfrac{dt}{dx}= -2x-4 \Rightarrow dt = ( -2x-4 )dx \)
Thus,
\( I_1 = \dfrac{-1}{2} \int \dfrac{-2x-4 }{\sqrt{-x^2-4x+5}}.dx \)
\( \Rightarrow I_1 = \dfrac{-1}{2} \int \dfrac{1}{\sqrt{t}}.dt \)
\( \Rightarrow I_1 = \dfrac{-1}{2} \int (t)^{\dfrac{-1}{2}}.dt \)
\( \Rightarrow I_1 =\dfrac{-1}{2} \left [ \dfrac{t^{\dfrac{-1}{2}+1}}{\left ( \dfrac{-1}{2}+1 \right )} \right ]+ C_1 \)
\( \Rightarrow I_1 =\dfrac{-1}{2} \left [ \dfrac{t^{\dfrac{1}{2}}}{\left ( \dfrac{1}{2} \right )} \right ]+ C_1 \)
\( \Rightarrow I_1 =- t^{\dfrac{1}{2}}+ C_1 \)
\( \Rightarrow I_1 =- \sqrt{t}+ C_1 \)
Putting back \( t = -x^2 -4x+5\)
\( \Rightarrow I_1 =- \sqrt{5-4x-x^2}+ C_1 \) ....(3)
Solving \( I_2\)
\( I_2 = \int \dfrac{1}{\sqrt{5-4x-x^2}}.dx \)
\( \Rightarrow I_2 = \int \dfrac{1}{\sqrt{5-4x-x^2}}.dx \)
\( \Rightarrow I_2 = \int \dfrac{1}{\sqrt{5-(x^2+4x)}}.dx \)
\( \Rightarrow I_2 = \int \dfrac{1}{\sqrt{5-(x^2+4x+4-4)}}.dx \)
\( \Rightarrow I_2 = \int \dfrac{1}{\sqrt{5-(x+2)^2+4}}.dx \)
\( \Rightarrow I_2 = \int \dfrac{1}{\sqrt{9-(x+2)^2}}.dx \)
\( \Rightarrow I_2 = \int \dfrac{1}{\sqrt{3^2-(x+2)^2}}.dx \)
\( \Rightarrow I_2 = \sin^{-1} \left ( \dfrac{x+2}{3} \right ) + C_2 \) ....(4)
\( \left [ \because \int \dfrac{1}{\sqrt{a^2-x^2}}.dx= \sin^{-1}\dfrac{x}{a} +C \right ] \)
Now, putting the value of \( I_1\) and \(I_2\) in equation \((2)\)
\( \int \dfrac{x+3}{\sqrt{5-4x-x^2} }.dx\)
\( = \dfrac{-1}{2} \int \dfrac{-2x-4}{\sqrt{5-4x-x^2} }.dx+ \int \dfrac{1}{\sqrt{5-4x-x^2} }.dx \)
\( = - \sqrt{5-4x-x^2} + C_1 + \sin^{-1} \left( \dfrac{x+2}{3}\right) + C_2 \)
\( = - \sqrt{5-4x-x^2} + C_1 + \sin^{-1} \left( \dfrac{x+2}{3}\right) + C \)
Where \(C = C_1+C_2\)
We can write numerator as
\( \Rightarrow x+2 = A \dfrac{d}{dx} ( 2x^2 +6x+5)+B ...(1)\)
\( \Rightarrow x+2 = A(4x+6)+B\)
\( \Rightarrow x+2 = 4Ax+(6A+B)\)
On comparing coefficient of \(x\) , we get
\(1= 4A \Rightarrow A = \dfrac{1}{4} \)
Comparing constant term
\( \Rightarrow 2 = 6A + B \Rightarrow \dfrac{6}{4} + B =2 \)
\( \Rightarrow B = 2- \dfrac{3}{2} = \dfrac{1}{2} \)
So,
\( A = \dfrac{1}{4} \) and \( B = \dfrac{1}{2} \)
Putting the value of \(A\) and \(B\) in \((1)\) , we get
\( \Rightarrow x+2 = \dfrac{1}{4}(4x+6) + \dfrac{1}{2} \)
Now, \( \int \dfrac{x+2}{2x^2 +6x+5}.dx = \int \dfrac{\dfrac{1}{4} (4x+6) + \dfrac{1}{2}}{2x^2 +6x+5}.dx = \dfrac{1}{4} \int \dfrac{4x+6}{2x^2 +6x+5\over{I_1}}.dx +\dfrac{1}{2} \int \dfrac{1}{2x^2 +6x+5\over{I_2}}.dx \) ....(2)
Solving \(I_1\)
\( \Rightarrow I_1= \dfrac{1}{4} \int \dfrac{4x+6}{2x^2 +6x+5}.dx \)
Taking \( t= 2x^2 +6x+5 \)
Differentiating both sides w.r.t.\(x\)
\( \Rightarrow \dfrac{dt}{dx}= (4x+6) \Rightarrow dt = (4x+6)dx \)
Thus,
\( I_1 = \int \dfrac{4x+6}{2x^2 +6x+5}.dx \)
\( \Rightarrow I_1 = \dfrac{1}{4} \int \dfrac{1}{t}.dt \)
\( \Rightarrow I_1 = \dfrac{1}{4} \text{log} |t| + C_1\)
\( \Rightarrow I_1 = \dfrac{1}{4} \text{log} |2x^2 +6x+5 | + C_1\)
Solving \( I_2\)
\( I_2 = \dfrac{1}{2} \int \dfrac{1}{2x^2 +6x+5}.dx \)
\( \Rightarrow I_2 = \dfrac{1}{2} \int \dfrac{1}{2\left [ x^2 + \dfrac{6x}{2} + \dfrac{5}{2}\right ]}.dx \)
\( \Rightarrow I_2 = \dfrac{1}{4} \int \dfrac{1}{x^2 +3x + \dfrac{5}{2} }.dx \)
\( \Rightarrow I_2 = \dfrac{1}{4} \int \dfrac{1}{x^2 +3x +\left ( \dfrac{3}{2} \right )^2 - \left ( \dfrac{3}{2} \right )^2+ \dfrac{5}{2} }.dx \)
\( \Rightarrow I_2 = \dfrac{1}{4} \int \dfrac{1}{\left ( x+ \dfrac{3}{2} \right )^2 -\dfrac{9}{4} + \dfrac{5}{2} }.dx \)
\( \Rightarrow I_2 = \dfrac{1}{4} \int \dfrac{1}{\left ( x+ \dfrac{3}{2} \right )^2 + \dfrac{1}{4} }.dx \)
\( \Rightarrow I_2 = \dfrac{1}{4} \int \dfrac{1}{\left ( x+ \dfrac{3}{2} \right )^2 + \left ( \dfrac{1}{2} \right )^2 }.dx \)
\( \Rightarrow I_2 = \dfrac{1}{4}\left [ \dfrac{1}{\dfrac{1}{2}} \tan^{-1} \left ( \dfrac{x+ \dfrac{3}{2}}{\dfrac{1}{2}} \right ) \right ] + C_2 \)
\( \left [ \because \int \dfrac{dx}{x^2+a^2}= \dfrac{1}{a} \tan^{-1}\dfrac{x}{a} +C \right ] \)
\( \Rightarrow I_2 = \dfrac{1}{4} [ 2~\tan^{-1} (2x+3)]+C_2\)
\( \Rightarrow I_2 = \dfrac{1}{2} \tan^{-1} (2x+3)+C_2\)
Now, putting the value of \( I_1\) and \(I_2\) in equation \((2)\)
\( \therefore \int \dfrac{x+2}{2x^2 +6x+5 }.dx\)
\( = \dfrac{1}{4} \text{log} |2x^2 +6x+5 | + C_1 +\dfrac{1}{2} \tan^{-1} (2x+3)+C_2 \)
\( = \dfrac{1}{4} \text{log} |2x^2 +6x+5 | +\dfrac{1}{2} \tan^{-1} (2x+3)+ C \)
(ii)\( \int \dfrac{x+3}{\sqrt{5-4x-x^2}}.dx\)
We can write numerator as
\( \Rightarrow x+3 = A \dfrac{d}{dx} ( -x^2 -4x+5)+B ...(1)\)
\( \Rightarrow x+3 = A(-2x-4)+B\)
\( \Rightarrow x+3 = -2Ax-4A+B\)
On comparing coefficient of \(x\) , we get
\( \Rightarrow 1= -2A \Rightarrow A = \dfrac{-1}{2} \)
Comparing constant term
\( \Rightarrow 3 = -4A + B \)
\( \Rightarrow B = -4 \left ( \dfrac{-1}{2} \right )+B\)
\( \Rightarrow B = 3-2=1\)
So,
\( A = \dfrac{-1}{2} \) and \( B = 1 \)
Putting the value of \(A\) and \(B\) in \((1)\) , we get
\( \Rightarrow x+3 = \dfrac{-1}{2}(-2x-4) +1 \)
Now, \( \int \dfrac{x+3}{\sqrt{5-4x-x^2}}.dx = \int \dfrac{\dfrac{-1}{2} (-2x-4) + 1}{\sqrt{5-4x-x^2} }.dx = \dfrac{-1}{4} \int \dfrac{-2x-6}{\sqrt{5-4x-x^2} \over{I_1}}.dx + \int \dfrac{1}{\sqrt{5-4x-x^2} \over{I_2}}.dx \) ....(2)
Solving \(I_1\)
\( \Rightarrow I_1= \dfrac{-1}{2} \int \dfrac{-2x-4}{ \sqrt{-x^2-4x+5}}.dx \)
Let \( t= -x^2 -4x+5\)
Differentiating both sides w.r.t.\(x\)
\( \Rightarrow \dfrac{dt}{dx}= -2x-4 \Rightarrow dt = ( -2x-4 )dx \)
Thus,
\( I_1 = \dfrac{-1}{2} \int \dfrac{-2x-4 }{\sqrt{-x^2-4x+5}}.dx \)
\( \Rightarrow I_1 = \dfrac{-1}{2} \int \dfrac{1}{\sqrt{t}}.dt \)
\( \Rightarrow I_1 = \dfrac{-1}{2} \int (t)^{\dfrac{-1}{2}}.dt \)
\( \Rightarrow I_1 =\dfrac{-1}{2} \left [ \dfrac{t^{\dfrac{-1}{2}+1}}{\left ( \dfrac{-1}{2}+1 \right )} \right ]+ C_1 \)
\( \Rightarrow I_1 =\dfrac{-1}{2} \left [ \dfrac{t^{\dfrac{1}{2}}}{\left ( \dfrac{1}{2} \right )} \right ]+ C_1 \)
\( \Rightarrow I_1 =- t^{\dfrac{1}{2}}+ C_1 \)
\( \Rightarrow I_1 =- \sqrt{t}+ C_1 \)
Putting back \( t = -x^2 -4x+5\)
\( \Rightarrow I_1 =- \sqrt{5-4x-x^2}+ C_1 \) ....(3)
Solving \( I_2\)
\( I_2 = \int \dfrac{1}{\sqrt{5-4x-x^2}}.dx \)
\( \Rightarrow I_2 = \int \dfrac{1}{\sqrt{5-4x-x^2}}.dx \)
\( \Rightarrow I_2 = \int \dfrac{1}{\sqrt{5-(x^2+4x)}}.dx \)
\( \Rightarrow I_2 = \int \dfrac{1}{\sqrt{5-(x^2+4x+4-4)}}.dx \)
\( \Rightarrow I_2 = \int \dfrac{1}{\sqrt{5-(x+2)^2+4}}.dx \)
\( \Rightarrow I_2 = \int \dfrac{1}{\sqrt{9-(x+2)^2}}.dx \)
\( \Rightarrow I_2 = \int \dfrac{1}{\sqrt{3^2-(x+2)^2}}.dx \)
\( \Rightarrow I_2 = \sin^{-1} \left ( \dfrac{x+2}{3} \right ) + C_2 \) ....(4)
\( \left [ \because \int \dfrac{1}{\sqrt{a^2-x^2}}.dx= \sin^{-1}\dfrac{x}{a} +C \right ] \)
Now, putting the value of \( I_1\) and \(I_2\) in equation \((2)\)
\( \int \dfrac{x+3}{\sqrt{5-4x-x^2} }.dx\)
\( = \dfrac{-1}{2} \int \dfrac{-2x-4}{\sqrt{5-4x-x^2} }.dx+ \int \dfrac{1}{\sqrt{5-4x-x^2} }.dx \)
\( = - \sqrt{5-4x-x^2} + C_1 + \sin^{-1} \left( \dfrac{x+2}{3}\right) + C_2 \)
\( = - \sqrt{5-4x-x^2} + C_1 + \sin^{-1} \left( \dfrac{x+2}{3}\right) + C \)
Where \(C = C_1+C_2\)
