What is the integral of (t^2)/2+(t^3)/3+(t^4)/4 ?

The integral of (t^2)/2+(t^3)/3+(t^4)/4 is (t^3)/6+(t^4)/(12)+(t^5)/(20)+C

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integral of (t^2)/2+(t^3)/3+(t^4)/4

integral

\frac{t ^{2}}{2} + \frac{t ^{3}}{3} + \frac{t ^{4}}{4}

\frac{t ^{3}}{6} + \frac{t ^{4}}{12} + \frac{t ^{5}}{20} + C

Solution steps

\int \frac{t ^{2}}{2} + \frac{t ^{3}}{3} + \frac{t ^{4}}{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 \frac{t ^{2}}{2} d t + \int \frac{t ^{3}}{3} d t + \int \frac{t ^{4}}{4} d t

\int \frac{t ^{2}}{2} d t = \frac{t ^{3}}{6}

\int \frac{t ^{3}}{3} d t = \frac{t ^{4}}{12}

\int \frac{t ^{4}}{4} d t = \frac{t ^{5}}{20}

= \frac{t ^{3}}{6} + \frac{t ^{4}}{12} + \frac{t ^{5}}{20}

Add a constant to the solution

= \frac{t ^{3}}{6} + \frac{t ^{4}}{12} + \frac{t ^{5}}{20} + C

(\frac{sin ( x )}{x})^{2}

sinh (5 x + 2)

\frac{- 4}{( x ^{2} + 4 ) ^{2}}

\frac{3 x ^{2} + 4}{( x ^{3} + 4 x ) ^{4}}

\frac{3}{( u ^{2} - 2 u )}

What is the integral of (t^2)/2+(t^3)/3+(t^4)/4 ?
The integral of (t^2)/2+(t^3)/3+(t^4)/4 is (t^3)/6+(t^4)/(12)+(t^5)/(20)+C