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Question Number 164809 by mathlove last updated on 22/Jan/22
1)∫x3x+x4=?2)∫x(x2+2x+2)2=?
Answered by Ar Brandon last updated on 22/Jan/22
Ostrogradsky∫x(x2+2x+2)2dx=px+qx2+2x+2+∫rx+sx2+2x+2dx⇒x(x2+2x+2)2=p(x2+2x+2)−(px+q)(2x+2)(x2+2x+2)2+rx+sx2+2x+2{r=x30p−2p+2r+s=x202p−2p−2q+2r+2s=x12p−2q+2s=x00⇒p=−12,q=−1,s=−12⇒∫x(x2+2x+2)2dx=−x+22(x2+2x+2)−12∫dxx2+2x+2⇒∫x(x2+2x+2)2dx=−x+22(x2+2x+2)−12tan−1(x+1)+C
I=∫x3x+x4dx,x=t12⇒dx=12t11dt=∫t4t6+t3(12t11dt)=12∫t12t2+1dt=12∫(t10−t8+t6−t4+t2−1+1t2+1)dt=12(t1111−t99+t77−t55+t33−t+tan−1(t))+C
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