Question

$∫^1_0 \frac{5x^2}{x^2+4x+3}dx$

Answer 1

I=$∫^1_0 \frac{5x^2}{x^2+4x+3}dx$

$Dividindg 5x^2 by x^2+4x+3,we obtain$

$I=∫^2_1{5-\frac{20x+15}{x^2+4x+3}}dx$

$=∫^2_1 5dx-∫^2_1\frac{20x+15}{x2+4x+3}dx$

$=[5x]^2_1-∫^2_1\frac{20x+15}{x^2+4x+3} dx$

$I=5-I1,where I=∫^2_1\frac{20x+15}{x2+4x+3}...(1)$

$Consider,I1=∫^2_1\frac{20x+15}{x^2+4x+8}dx$

$Let 20x+15=A\frac{d}{dx}(x^2+4x+3)+B$

$=2Ax+(4A+B)$

Equating the coefficients of x and constant term,we obtain

$A=10 and B=-25$

$⇒I1=10 ∫^2_1\frac{ 2x+4}{x2+4x+3}dx-25 ∫^2_1{dx}{x^2+4x+3}$

$Let x^2+4x+3=t$

$⇒(2x+4)dx=dt$

$⇒I_1=10∫\frac{dt}{t}-25∫\frac{dx}{(x+2)^2-1^2}$

$=10log t-25[\frac{1}{2}log(\frac{x+2-1}{x+2+1})]$

$=[10log(x2+4x+3)]21-25[\frac{1}{2}log\frac{x+1}{x+3)}]^2_1$

$=[10log5+10log3-10log4-10log2]-\frac{25}{2}[log3-log5-log2+log4]$

$=[10+\frac{25}{2}]log5+[-10-\frac{25}{2}]log4+[10-\frac{25}{2}]log3+[-10+\frac{25}{2}]log2$

$=\frac{45}{2}log5-\frac{45}{2}log4-\frac{5}{2}log3+\frac{5}{2}log2$

$=\frac{45}{2}log\frac{5}{4}-\frac{5}{2}log\frac{3}{2}$

$Substituting the value of I1 in(1),we obtain$

$I=5-[\frac{45}{2}log\frac{5}{4}-\frac{5}{2}log\frac{3}{2]}$

$=5-\frac{5}{2}[9log\frac{5}{4}-log\frac{3}{2}]$