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

$$find\int (v^3-2)/(v^4+v)dv$$

Answered by mhmd last updated on 23/Aug/19

Answered by MJS last updated on 23/Aug/19

$$\int \frac{v^3-2}{v^4+v}dv=\int \frac{4v^3+1}{4(v^4+v)}dv-\int \frac{9}{4(v^4+v)}dv=$$$$$$$$\int \frac{4v^3+1}{4(v^4+v)}dv=\frac{1}{4}\int \frac{4v^3+1}{v^4+v}dv=$$$$[t=v^4+v\to dv=\frac{dt}{4v^3+1}]$$$$=\frac{1}{4}\int \frac{dt}{t}=\frac{1}{4}\ln t=\frac{1}{4}\ln (v^4+v)=\frac{1}{4}\ln v+\frac{1}{4}\ln (v^3+1)$$$$$$$$-\int \frac{9}{4(v^4+v)}dv=-\frac{9}{4}\int \frac{dv}{v^4+v}=-\frac{9}{4}\int \frac{dv}{v^4(1+\frac{1}{v^3})}=$$$$[u=1+\frac{1}{v^3}\to dv=-\frac{v^4}{3}]$$$$=\frac{3}{4}\int \frac{du}{u}=\frac{3}{4}\ln u=\frac{3}{4}\ln (1+\frac{1}{v^3})=\frac{3}{4}\ln (v^3+1)-\frac{9}{4}\ln v$$$$$$$$=\ln (v^3+1)-2\ln v+C$$

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