find-v-3-2-v-4-v-dv-

Question Number 67153 by mhmd last updated on 23/Aug/19

f i n d \int (v^{3} - 2) / (v^{4} + v) d v

Answered by mhmd last updated on 23/Aug/19

Answered by MJS last updated on 23/Aug/19

\int \frac{v^{3} - 2}{v^{4} + v} d v = \int \frac{4 v^{3} + 1}{4 (v^{4} + v)} d v - \int \frac{9}{4 (v^{4} + v)} d v =

\int \frac{4 v^{3} + 1}{4 (v^{4} + v)} d v = \frac{1}{4} \int \frac{4 v^{3} + 1}{v^{4} + v} d v =

[t = v^{4} + v \to d v = \frac{d t}{4 v^{3} + 1}]

= \frac{1}{4} \int \frac{d t}{t} = \frac{1}{4} \ln t = \frac{1}{4} \ln (v^{4} + v) = \frac{1}{4} \ln v + \frac{1}{4} \ln (v^{3} + 1)

- \int \frac{9}{4 (v^{4} + v)} d v = - \frac{9}{4} \int \frac{d v}{v^{4} + v} = - \frac{9}{4} \int \frac{d v}{v^{4} (1 + \frac{1}{v^{3}})} =

[u = 1 + \frac{1}{v^{3}} \to d v = - \frac{v^{4}}{3}]

= \frac{3}{4} \int \frac{d u}{u} = \frac{3}{4} \ln u = \frac{3}{4} \ln (1 + \frac{1}{v^{3}}) = \frac{3}{4} \ln (v^{3} + 1) - \frac{9}{4} \ln v

= \ln (v^{3} + 1) - 2 \ln v + C

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