What is the integral of ((t^3)/3+1/(4t^2)) ?

The integral of ((t^3)/3+1/(4t^2)) is (t^4)/(12)-1/(4t)+C

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integral of ((t^3)/3+1/(4t^2))

\int (\frac{t ^{3}}{3} + \frac{1}{4 t ^{2}}) d t

\frac{t ^{4}}{12} - \frac{1}{4 t} + C

Solution steps

\int \frac{t ^{3}}{3} + \frac{1}{4 t ^{2}} 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 ^{3}}{3} d t + \int \frac{1}{4 t ^{2}} d t

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

\int \frac{1}{4 t ^{2}} d t = - \frac{1}{4 t}

= \frac{t ^{4}}{12} - \frac{1}{4 t}

Add a constant to the solution

= \frac{t ^{4}}{12} - \frac{1}{4 t} + C

\frac{\partial}{\partial x} (- 3 sin (4 x))

t - 4

f (x) = \frac{7}{x , at ( 4 , \frac{7}{2} )}

f (x) = - x^{3}

x^{2} \frac{d y}{d x} + (x^{2} + 2 x) y = e^{x}

What is the integral of ((t^3)/3+1/(4t^2)) ?
The integral of ((t^3)/3+1/(4t^2)) is (t^4)/(12)-1/(4t)+C