Solve ∫ t^4+t^2+1/t^2+t+1dt

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

Evaluate

$$\frac{t^{3}}{3}-\frac{t^{2}}{2}+t+С$$

Solution Steps

Factor the expressions that are not already factored in \frac{t^{4}+t^{2}+1}{t^{2}+t+1}.

$$\int \frac{\left(t^{2}+t+1\right)\left(t^{2}-t+1\right)}{t^{2}+t+1}\mathrm{d}t$$

Cancel out t^{2}+t+1 in both numerator and denominator.

$$\int t^{2}-t+1\mathrm{d}t$$

Integrate the sum term by term.

$$\int t^{2}\mathrm{d}t+\int -t\mathrm{d}t+\int 1\mathrm{d}t$$

Factor out the constant in each of the terms.

$$\int t^{2}\mathrm{d}t-\int t\mathrm{d}t+\int 1\mathrm{d}t$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{2}\mathrm{d}t with \frac{t^{3}}{3}.

$$\frac{t^{3}}{3}-\int t\mathrm{d}t+\int 1\mathrm{d}t$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t\mathrm{d}t with \frac{t^{2}}{2}. Multiply -1 times \frac{t^{2}}{2}.

$$\frac{t^{3}}{3}-\frac{t^{2}}{2}+\int 1\mathrm{d}t$$

Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}t=at.

$$\frac{t^{3}}{3}-\frac{t^{2}}{2}+t$$

Simplify.

$$t-\frac{t^{2}}{2}+\frac{t^{3}}{3}$$

If F\left(t\right) is an antiderivative of f\left(t\right), then the set of all antiderivatives of f\left(t\right) is given by F\left(t\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$t-\frac{t^{2}}{2}+\frac{t^{3}}{3}+С$$

Differentiate w.r.t. t

$$t^{2}-t+1$$