Find : ∫(2x−5)e2x(2x−3)3dx. OR Find : ∫(x2+x+1)(x2+1)(x+2)dx.
Let I=∫(2x−5)e2x(2x−3)3dx Put 2x=t⇒dx=12dt
∴I=12∫(t−5)et(t−3)3dt ⇒2I=∫[(t−3)−2(t−3)3]etdt⇒2I=∫[1(t−3)2+−2(t−3)3]etdt ⇒2I=∫1(t−3)2×etdt−∫[2(t−3)3]etdt⇒2I=1(t−3)2×∫etdt−∫(ddt1(t−3)2∫etdt)dt−∫[2(t−3)3]etdt⇒2I=1(t−3)2×et−∫(−2(t−3)2×et)dt−∫[2(t−3)3]etdt=1(t−3)2×et+C∴I=12(2x−3)2× e2x+C.
OR
I=∫x2+x+1(x2+1)(x+2)dxConsider x2+x+1(x2+1)(x+2)=Ax+Bx2+1+Cx+2 ⇒x2+x+1=(Ax+B)(x+2)+C(x2+1)On comparing like terms, we get:A=25,B=15,C=35∴I=15∫2x(x2+1)dx+15∫1(x2+1)dx+35∫1(x+2)dx⇒1=15log|x2+1|+15tan−1x+35log|x+2|.