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Evaluate the definite integral\[\int\limits_1^2\frac{5x^2}{x^2+4x+3}\]

$\begin{array}{1 1} 5- \large \frac{5}{2}[9 \log(5/4)-\log(3/2)] \\ 5- \large \frac{5}{2}[7 \log(5/4)+\log(3/2)] \\5+ \large \frac{5}{2}[9 \log(5/4)+\log(3/2)]\\ 5- \large \frac{3}{2}[7 \log(5/4)-\log(5/2)] \end{array} $

1 Answer

  • (i)$\int \limits_a^bf(x)dx=F(b)-F(a)$
  • (ii)If the given rational function is improper in natture,then we can divide and split it into a proper rational function and then resolve it into partial fraction of the given form.
  • (iii)$\int \frac{dx}{(a+x)}=log(a+x)+c$
Given $\int\limits_1^2\large\frac{5x^2}{x^2+4x+3}$
The given function is an improper rational function. Hence on dividing we get
$\large\frac{20x+15}{x^2+4x+3}$ can be resolved as $ \large\frac{A}{(x+3)}+\frac{B}{(x+1)}$
Became $(x^2+4x+3)$ can be factorised as $(x+3)(x+1)$
Therefore $20x+15=A(x+1)+B(x+3)$
Equating the coefficient of x,
$ 20=A+B\;------(1)$
Equating the constant term,
On solving equation (1) and(2),
$\qquad 2B=-5$
Therefore $B=-\frac{5}{2}$
Substituting the value of B in equ(1)
Hence $A=\frac{45}{2}\;and\;B=-\frac{5}{2}$
Now substituting for A and B we get
On integtating we get
$\int \limits_1^2 \large\frac{5x^2}{x^2+4x+3}=\int \limits_1^2 5dx -\bigg[\int \limits_1^2 \frac{45}{2}\bigg(\frac{1}{x+3}\bigg)dx-\int
\limits_1^2 \frac{5}{2}\bigg(\frac{1}{x+1}\bigg)dx\bigg]$
$=\bigg[5x\bigg]_1^2-\large\frac{45}{2}[log(x+3)]_1^2+\frac{5}{2}[log (x+1)]_1^2$
On applying limits,
$[5 \times 2-5 \times 1]-\frac{45}{2}[log(x+3)-log(1+3)]-\frac{5}{2}[log(2+1)-log(1+1)]$
But we know $log(a)-log(b)=log(a/b)$
Therefore $ \int \limits_1^2 \large\frac{5x^2}{x^2+4x+3}=5-\frac{45}{2}[log(5/4)-log(3/2)]$
$ \int \limits_1^2 \large\frac{5x^2}{x^2+4x+3}=5-\frac{5}{2}[9log(5/4)-log(3/2)]$


answered Feb 11, 2013 by meena.p