Solve ∫ sqrt[5]{t^4}+t^2(10t^3-5t)dt

$$\int \sqrt[ 5 ] { t ^ { 4 } } + t ^ { 2 } ( 10 t ^ { 3 } - 5 t ) d t$$

Evaluate

$$\frac{5t^{6}}{3}-\frac{5t^{4}}{4}+\frac{5t^{\frac{9}{5}}}{9}+С$$

Solution Steps

Use the distributive property to multiply t^{2} by 10t^{3}-5t.

$$\int \sqrt[5]{t^{4}}+10t^{5}-5t^{3}\mathrm{d}t$$

Integrate the sum term by term.

$$\int t^{\frac{4}{5}}\mathrm{d}t+\int 10t^{5}\mathrm{d}t+\int -5t^{3}\mathrm{d}t$$

Factor out the constant in each of the terms.

$$\int t^{\frac{4}{5}}\mathrm{d}t+10\int t^{5}\mathrm{d}t-5\int t^{3}\mathrm{d}t$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{\frac{4}{5}}\mathrm{d}t with \frac{5t^{\frac{9}{5}}}{9}.

$$\frac{5t^{\frac{9}{5}}}{9}+10\int t^{5}\mathrm{d}t-5\int t^{3}\mathrm{d}t$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{5}\mathrm{d}t with \frac{t^{6}}{6}. Multiply 10 times \frac{t^{6}}{6}.

$$\frac{5t^{\frac{9}{5}}}{9}+\frac{5t^{6}}{3}-5\int t^{3}\mathrm{d}t$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{3}\mathrm{d}t with \frac{t^{4}}{4}. Multiply -5 times \frac{t^{4}}{4}.

$$\frac{5t^{\frac{9}{5}}}{9}+\frac{5t^{6}}{3}-\frac{5t^{4}}{4}$$

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.

$$\frac{5t^{\frac{9}{5}}}{9}+\frac{5t^{6}}{3}-\frac{5t^{4}}{4}+С$$

Differentiate w.r.t. t

$$10t^{5}-5t^{3}+t^{\frac{4}{5}}$$