Evaluate the definite integral _ 1 ^ 2 5 x ^ 2 x ^ 2 + 4 x + 3

Question

Evaluate the definite integral $ \int _ { 1 } ^ { 2 } \frac { 5 x ^ { 2 } } { x ^ { 2 } + 4 x + 3 }$

✓

Answer

According to the question, $ I = \int _ { 1 } ^ { 2 } \frac { 5 x ^ { 2 } } { x ^ { 2 } + 4 x + 3 } d x$
$ I = 5 \int _ { 1 } ^ { 2 } \left( 1 + \frac { - 4 x - 3 } { x ^ { 2 } + 4 x + 3 } \right) d x$
$ = 5 \int _ { 1 } ^ { 2 } d x - 5 \int _ { 1 } ^ { 2 } \frac { 4 x + 3 } { x ^ { 2 } + 4 x + 3 } d x$
$ \Rightarrow \quad I = 5 [ x ] _ { 1 } ^ { 2 } - 5 \int _ { 1 } ^ { 2 } \frac { 4 x + 3 } { ( x + 3 ) ( x + 1 ) } d x$
Using partial fraction,
let $ \frac { 4 x + 3 } { ( x + 3 ) ( x + 1 ) } = \frac { A } { x + 3 } + \frac { B } { x + 1 }$
$ \Rightarrow \quad \frac { 4 x + 3 } { ( x + 3 ) ( x + 1 ) } = \frac { A ( x + 1 ) + B ( x + 3 ) } { ( x + 3 ) ( x + 1 ) }$
$ \Rightarrow$ $4x + 3 = A(x + 1) + B(x + 3)$
Comparing the coefficients of like from both sides,
$\Rightarrow A + B = 4 \Rightarrow A = 4 - B$
and A + 3B = 3$ \Rightarrow$ $4 - B + 3B = 3$
$ \Rightarrow \quad B = - \frac { 1 } { 2 },$then $ A = 4 + \frac { 1 } { 2 } = \frac { 9 } { 2 }$
Now, from Equation (i), we get
$ I = 5 ( 2 - 1 ) - 5 \int _ { 1 } ^ { 2 } \left( \frac { 9 / 2 } { x + 3 } + \frac { - 1 / 2 } { x + 1 } \right) d x$
$ = 5 - 5 \left[ \frac { 9 } { 2 } \log | x + 3 | - \frac { 1 } { 2 } \log | x + 1 | \right] _ { 1 } ^ { 2 }$
$ = 5 - 5\left. {\left[ {\left( {\frac{9}{2}\log 5 - \frac{1}{2}\log 3} \right)} \right. - \left( {\frac{9}{2}\log 4 - \frac{1}{2}\log 2} \right)} \right]$
$ = 5 - 5 \left[ \frac { 9 } { 2 } ( \log 5 - \log 4 ) - \frac { 1 } { 2 } ( \log 3 - \log 4 ]\right.$
$ = 5 - 5 \left[ \frac { 9 } { 2 } \log \frac { 5 } { 4 } - \frac { 1 } { 2 } \log \frac { 3 } { 2 } \right]$
$ = 5 - \frac { 45 } { 2 } \log \frac { 5 } { 4 } + \frac { 5 } { 2 } \log \frac { 3 } { 2 }$