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

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

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

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

t - \frac{4}{3 t ^{3}} + C

Solution steps

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

Expand

\frac{t ^{4} + 4}{t ^{4}} : 1 + \frac{4}{t ^{4}}

= \int 1 + \frac{4}{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 1 d t + \int \frac{4}{t ^{4}} d t

\int 1 d t = t

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

= t - \frac{4}{3 t ^{3}}

Add a constant to the solution

= t - \frac{4}{3 t ^{3}} + C

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