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Question Number 175614 by leodera last updated on 03/Sep/22

$$\int \frac{1}{4t^3+3t^2+4t+1}dt$$

Answered by ajfour last updated on 04/Sep/22

$$I=\int \frac{dt}{4(t+p)(t^2+qt+r)}$$$$4I=\int \frac{Adt}{t+p}+\frac{(Bt+C)dt}{t^2+qt+r}$$$$4I=A\ln (t+p)+\frac{B}{2}\ln (t^2+qt+r)$$$$+s{\tan }^{-1}\frac{t+\frac{q}{2}}{\sqrt{r-\frac{q^2}{4}}}+c$$$$differentiating$$$$\frac{A}{t+p}+\frac{B(t+\frac{q}{2})}{t^2+qt+r}+\frac{s\sqrt{r-\frac{q^2}{4}}}{{(t+\frac{q}{2})}^2+r-\frac{q^2}{4}}$$$$\frac{A(t^2+qt+r)+B(t+\frac{q}{2})(t+p)+s\sqrt{r-\frac{q^2}{4}}}{(t+p)(t^2+qt+r)}$$$$=\frac{1}{t^3+\frac{3}{4}{t}^2+t+\frac{1}{4}}$$$$comparetofind$$$$p,q,r,A,B,s$$$$$$

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