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Question Number 45373 by arvinddayama01@gmail.com last updated on 12/Oct/18

$$\int \frac{{\mathrm{t}}^3}{1+\mathrm{t}}\mathrm{dt}=?$$

Commented by maxmathsup by imad last updated on 12/Oct/18

$$\int \frac{t^3}{1+t}dt=\int \frac{t^3+1-1}{1+t}dt=\int (t^2-t+1)dt-\int \frac{dt}{1+t}$$$$=\frac{t^3}{3}-\frac{t^2}{2}+t-ln\mid 1+t\mid +k.$$

Answered by ajfour last updated on 12/Oct/18

$$\int \frac{(1+t)(1-t+t^2)}{1+t}dt-\int \frac{dt}{1+t}$$$$=t-\frac{t^2}{2}+\frac{t^3}{3}-\ln \mid 1+t\mid +c.$$

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