What is the integral of (x^4+1)/(x^4) ?

The integral of (x^4+1)/(x^4) is x-1/(3x^3)+C

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integral of (x^4+1)/(x^4)

\int \frac{x ^{4} + 1}{x ^{4}} d x

x - \frac{1}{3 x ^{3}} + C

Solution steps

\int \frac{x ^{4} + 1}{x ^{4}} d x

Expand

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

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

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 x + \int \frac{1}{x ^{4}} d x

\int 1 d x = x

\int \frac{1}{x ^{4}} d x = - \frac{1}{3 x ^{3}}

= x - \frac{1}{3 x ^{3}}

Add a constant to the solution

= x - \frac{1}{3 x ^{3}} + C

\int \frac{1}{1 - x} d x

\int \frac{x}{( 2 + x ^{2} ) ^{4}} d x

\int (9 x^{2} + 2 x) sec^{2} (3 x^{3} + x^{2}) d x

\int (2 x^{3} x^{2} - 5) d x

\int \frac{1 + x}{x} d x

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