Question

Evaluate the definite integral: $∫_0^1\frac{5x^2}{x2+4x+3}$

Answer 1

Let I=$∫_0^1\frac{5x^2}{x^2+4x+3}$

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

$I=∫_1^2-5\frac{20x+15}{x^2+4x+3}dx$

$=∫_1^2 5dx-\int_1^2\frac{20x+15}{x^2+4x+3}dx$

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

$I=5-I_1,where I=∫_1^2\frac{20x+15}{x^2+4x+3}...(1)$

$Consider,I_1=∫_1^2\frac{20x+15}{x^2+4x+8}dx$

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

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

Equating the coefficients of x and constant term,we obtain

$A=10 and B=-25$

$⇒I_1=10 ∫_1^2\frac{2x+4}{x^2+4x+3}dx-25 ∫_1^21\frac{dx}{x^2+4x+3}$

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

$⇒(2x+4)dx=dt$

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

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

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

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

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

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

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

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

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