\(∫^1_0 \frac{5x^2}{x^2+4x+3}dx\)
Solution and Explanation
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
\(A=10 and B=-25\)
\(⇒I1=10 ∫^2_1\frac{ 2x+4}{x2+4x+3}dx-25 ∫^2_1{dx}{x^2+4x+3}\)
\(Let x^2+4x+3=t\)
\(⇒(2x+4)dx=dt\)
\(⇒I_1=10∫\frac{dt}{t}-25∫\frac{dx}{(x+2)^2-1^2}\)
\(=10log t-25[\frac{1}{2}log(\frac{x+2-1}{x+2+1})]\)
\(=[10log(x2+4x+3)]21-25[\frac{1}{2}log\frac{x+1}{x+3)}]^2_1\)
\(=[10log5+10log3-10log4-10log2]-\frac{25}{2}[log3-log5-log2+log4]\)
\(=[10+\frac{25}{2}]log5+[-10-\frac{25}{2}]log4+[10-\frac{25}{2}]log3+[-10+\frac{25}{2}]log2\)
\(=\frac{45}{2}log5-\frac{45}{2}log4-\frac{5}{2}log3+\frac{5}{2}log2\)
\(=\frac{45}{2}log\frac{5}{4}-\frac{5}{2}log\frac{3}{2}\)
\(Substituting the value of I1 in(1),we obtain\)
\(I=5-[\frac{45}{2}log\frac{5}{4}-\frac{5}{2}log\frac{3}{2]}\)
\(=5-\frac{5}{2}[9log\frac{5}{4}-log\frac{3}{2}]\)
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Concepts Used:
Fundamental Theorem of Calculus
Fundamental Theorem of Calculus is the theorem which states that differentiation and integration are opposite processes (or operations) of one another.
Calculus's fundamental theorem connects the notions of differentiating and integrating functions. The first portion of the theorem - the first fundamental theorem of calculus – asserts that by integrating f with a variable bound of integration, one of the antiderivatives (also known as an indefinite integral) of a function f, say F, can be derived. This implies the occurrence of antiderivatives for continuous functions.