Engineering Board Exam Reviewer · Integration by Substitution · medium

Evaluate: x3 x2 + 4,dx

1. ≤ft(x2 + 4 )5/25 + 4 ≤ft(x2 + 4 )3/23+C
2. ≤ft(x2 + 4 )5/25 - ≤ft(x2 + 4 )3/23+C
3. ≤ft(x2 + 4 )5/25 - 4 ≤ft(x2 + 4 )3/23+C
4. ≤ft(x2 + 4 )5/210 - 2 ≤ft(x2 + 4 )3/23+C

Answer

C. ≤ft(x2 + 4 )5/25 - 4 ≤ft(x2 + 4 )3/23+C

Solution

Let u=x2+4, so x2=u-4 and x,dx= 12du. Then x3sqrt(x2+4),dx= 12 (u-4)u1/2,du= 12≤ft( 25u5/2- 83u3/2 )= u5/25- 4u3/23+C with u=x2+4.

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