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=5−20x+15x2+4x+3∴I=∫21[5−20x+15x2+4x+3]dx=∫215dx−∫2120x+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=452∴I=5∫211dx+52∫211(x+1)dx−452∫211(x+3)dx[fromEq(i)]=[5x+52log|x+1|−452log|x+3|]21=[10+52log|3|−452log|5|−5−52log|2|+452log|4|]⇒I=5+52log∣∣32∣∣−452log∣∣54∣∣[∵mloga−mlogb=mlog(ab)] Mathematics

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