a) Ans:
Let "3x+1\\equiv \\alpha(\\text{derivative of }4x^2+12x+5)+\\beta"
"\\implies \\ 3x+1=\\alpha(8x+12)+\\beta"
"\\implies \\ 3x+1=8\\alpha x+(12\\alpha+\\beta)"
On equating the coefficients of constant term and variable x both side ,we get
"8\\alpha=3 \\ and \\ 12\\alpha+\\beta=1"
"\\alpha=\\frac{3}{8} \\ and \\ \\beta=1-12\\alpha"
"=1-12\u00d7\\frac{3}{8}"
"=\\frac{-7}{2}"
"\\therefore 3x+1=\\frac{3}{8}(8x+12)-\\frac{7}{2}"
"\\therefore I=\\frac{3}{8}\\intop (8x+12)\\sqrt{4x^2+12x+5} \\ -\\frac{7}{2}\\int\\sqrt{4x^2+12x+5}"
"=I_1-I_2 \\ (says)"
Where, "I_1=\\frac{3}{8}\\int(8x+12)\\sqrt{4x^2+12x+5} \\ and \\ I_2=\\frac{7}{2}\\int\\sqrt{4x^2+12x+5}"
Let "4x^2+12x+5=t \\ \\implies\\frac{dt}{dx}=8x+12"
"\\implies dt=(8x+12)dx" .
"=\\frac{3}{8}\u00d7\\frac{2}{3}\u00d7t^{\\frac{3}{2}}+c_1 \\ where \\ c_1" is a integration constant.
"=\\frac{1}{4} {(4x^2+12x+5)}^{\\frac{3}{2}} +c_1"
Now ,"I_2=\\frac{7}{2}\\int\\sqrt{4x^2+12x+5}=\\frac{7}{2}\\int\\sqrt{{(2x)}^2+2.2x.3+3^2-2^2}"
"=\\frac{7}{2}\\int\\sqrt{{(2x+3)}^2-2^2}"
Put "2x+3=v\\implies 2=\\frac{dv}{dt} \\implies dv=2dt"
"\\therefore I_2=\\frac{7}{2} \u00d7 2 \\int\\sqrt{v^2-2^2}"
Which is ,of the form "\\int\\sqrt{x^2-a^2}dx" ,therefore
"I_2=7 (\\frac{v\\sqrt{v^2-2^2}}{2} -\\frac{2^2}{2}log(v+\\sqrt{v^2-2^2}))+c_2"
Where "c_2" is an integration constant.
"=7[\\frac{(2x+3)\\sqrt{4x^2+12x+5}}{2}-2log(2x+3+\\sqrt{4x^2+12x+5} \\ ]+c_2"
Therefore,
"I=\\frac{1}{4}\u00d7{(4x^2+12x+5)}^\\frac{3}{2}-\\frac{7}{2}(2x+3)\\sqrt{4x^2+12x+5}+2log(2x+3+\\sqrt{4x^2+12x+5})+k"
"\\text{where}, k=c_1-c_2" is a integration constant.
b)Ans:
"=log(x+1)+tan^{-1}x+c"
Where c is an integration constant.
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