What is the integral of (t^2-2t^4) ?

The integral of (t^2-2t^4) is (t^3)/3-(2t^5)/5+C

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integral of (t^2-2t^4)

\int (t^{2} - 2 t^{4}) d t

\frac{t ^{3}}{3} - \frac{2 t ^{5}}{5} + C

Solution steps

\int t^{2} - 2 t^{4} d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int t^{2} d t - \int 2 t^{4} d t

\int t^{2} d t = \frac{t ^{3}}{3}

\int 2 t^{4} d t = \frac{2 t ^{5}}{5}

= \frac{t ^{3}}{3} - \frac{2 t ^{5}}{5}

Add a constant to the solution

= \frac{t ^{3}}{3} - \frac{2 t ^{5}}{5} + C

\int_{1}^{2} \frac{1}{x ^{5}} d x

\frac{\partial}{\partial x} (y e^{4 x})

\int_{1}^{5} \frac{ln ( x )}{x ^{7}} d x

\frac{d}{d x} (\frac{3}{2} x^{\frac{5}{3}} - \frac{5 x}{4} + 7 x^{2} - \frac{6}{x ^{5}} + 6)

y = (3 x - 2 x^{2})^{3} \frac{d y}{d x}

What is the integral of (t^2-2t^4) ?
The integral of (t^2-2t^4) is (t^3)/3-(2t^5)/5+C