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Question

Evaluate the definite integrals.
215x2x2+4x+3dx


Solution

Let I=215x2x2+4x+3dx
Here, the degree of numerator and denominator is same, so on dividing numerator by denominator, we get
5x2x2+4x+3=520x+15x2+4x+3I=21[520x+15x2+4x+3]dx=215dx2120x+15(x+1)(x+3).....(i)
Let 20x+15(x+1)(x+3)dx=A(x+1)+B(x+3)
20x+15=A(x+3)+B(x+1)20x+15=Ax+3A+Bx+B
On equating the coefficients of x and constant term on both sides, we get
20=A+B .....(ii)
and 15=3A +B .....(iii)
On solving Eqs.(ii)and (iii), we get
A=52,B=452I=5211dx+52211(x+1)dx452211(x+3)dx[fromEq(i)]=[5x+52log|x+1|452log|x+3|]21=[10+52log|3|452log|5|552log|2|+452log|4|]I=5+52log32452log54[mlogamlogb=mlog(ab)]


Mathematics

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