Answer to Question #107655 in Calculus for Parul

Question #107655
a) integrate (3x+1)√4x^2 + 12x + 5 dx
b) integrate x^2+x+5 / (x^2+4)(x+1)
1
Expert's answer
2020-04-13T12:28:49-0400

a) Ans:


"\\text{Let} \\ I=\\int ( 3x+1)\\sqrt{4x^2+12x+5} \\ dx"


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" .



"\\therefore I_1=\\frac{3}{8}\\int\\sqrt{t} dt \\ =\\frac{3}{8} \u00d7 \\frac{t^{1+\\frac{1}{2}}}{1+\\frac{1}{2}}+c_1"

"=\\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:


"\\int\\frac{x^2+x+5}{(x^2+4)(x+1)}=\\int\\frac{(x^2+4)+(x+1)}{(x^2+4)(x+1)}"




"=\\int\\frac{x^2+4}{(x^2+4)(x+1)}+\\int\\frac{x+1}{(x^2+1)(x+1)}"


"=\\int\\frac{1}{x+1}+\\int\\frac{1}{x^2+1}"



"=log(x+1)+tan^{-1}x+c"

Where c is an integration constant.



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS