Indefinite Integrals

Question

Hi, could someone walk me through this problem? 

 

Indefinite integral of [v + 1/((4v-7)^8)]

Answer 1

Hannah,

 

Just to be sure, the v+1 is NOT in the numerator of the fraction, right?

 

Assuming the answer is YES, here is a solution:

 

∫ [v + 1/(4v-7)⁸] dv = v²/2 + ∫ (4v-7)^-8 dv.

 

The remaining integral may be done by simpl3 substitution, using u = 4v - 7, du = 4 dv, dv = du/4:

 

Original integral = v²/2 + 1/4 ∫ u^-8 du = v²/2 + 1/4 u^-7/(-7) + C

 

    = v²/2 - 1/28 u^-7 + C

 

    = v²/2 - 1/28 (4v - 7)^-7 + C.

 

If needed, the second term can be written as a fraction: -1/[28(4v-7)⁷].