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Evaluate the definite integrals in Exercises 1 to 20.

Q16. $\int_1^2\frac{5x^2}{x^2 + 4x +3}$

Answers (1)

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Given integral: $I = \int_1^2\frac{5x^2}{x^2 + 4x +3}$

So, we can rewrite the integral as;

$I = \int_1^2 \frac{5x^2}{x^2 + 4x +3}= \int_1^2 \left ( 5 - \frac{20x+15}{x^2 + 4x +3} \right ) dx$

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

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

$I = 5-I_1$ where $I = \int_1^2 \frac{20x+15}{x^2+4x+3}dx$ .................(1)

Now, consider $I = \int_1^2 \frac{20x+15}{x^2+4x+3}dx$

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

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

We now equate the coefficients of x and constant term, we get

$A= 10 \and\ B =-25$

$\Rightarrow I_1 = 10\int_1^2 \frac{2x+4}{x^2+4x+3}dx -25\int_1^2 \frac{dx}{x^2+4x+3}$

Now take denominator $x^2+4x+3 = t$

Then we have $(2x+4)dx =dt$

$\Rightarrow I_{1} =10\int \frac{dt}{t} -25\int \frac{dx}{(x+2)^2-1^2}$

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

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

$= \left [ 10\log15 -10\log 8 \right ] -25 \left [ \frac{1}{2}\log\frac{3}{5} -\frac{1}{2}\log\frac{2}{4} \right ]$

$= \left [ 10\log5 +10\log3 -10\log4-10\log2 \right ] -\frac{25}{2}\left [ \log3 -\log5-\log2+\log4 \right ]$ $= \left ( 10+\frac{25}{2} \right )\log5 + \left ( -10-\frac{25}{2} \right )\log 4 + \left ( 10-\frac{25}{2} \right )\log 3 + \left ( -10+\frac{25}{2} \right )\log 2$ $= \frac{45}{2}\log5 -\frac{45}{2}\log4 - \frac{5}{2}\log3 +\frac{5}{2}\log2$

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

Then substituting the value of $I_{1}$ in equation (1), we get

$I= 5 -\left ( \frac{45}{2}\log\frac{5}{4}-\frac{5}{2}\log\frac{3}{2} \right )$

$= 5 -\frac{5}{2}\left ( 9\log\frac{5}{4}-\log\frac{3}{2} \right )$

Posted by

Divya Prakash Singh

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