# Calculate antiderivative y = 1/(x^3+x^4)?

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### 1 Answer

You need to evaluate the anti-derivative of the given function, hence, you need to evaluate the following indefinite integral, such that:

`int f(x)dx = int 1/(x^3+x^4) dx`

You need to perform the following operations, such that:

`1/(x^3+x^4) = 1/(x^3(1 + x))`

`1/(x^3(1 + x)) = (1 + x - x)/(x^3(1 + x))`

`(1 + x - x)/(x^3(1 + x)) = (1 + x)/(x^3(1 + x)) - x/(x^3(1 + x))`

Reducing duplicate factors yields:

`(1 + x - x)/(x^3(1 + x)) = 1/x^3 - 1/(x^2(1 + x))`

You need to perform the same operations in case of the fraction `1/(x^2(1 + x))` , such that:

`1/(x^2(1 + x)) = (1 + x - x)/(x^2(1 + x)) `

`(1 + x - x)/(x^2(1 + x)) = (1 + x)/(x^2(1 + x)) - x/(x^2(1 + x))`

Reducing duplicate factors yields:

`(1 + x - x)/(x^2(1 + x)) = 1/x^2 - 1/(x(1+x))`

You need to perform the same operations in case of the fraction `1/(x(1 + x))` , such that:

`1/(x(1 + x)) = (1 + x - x)/(x(1 + x)) `

`(1 + x - x)/(x(1 + x)) = (1 + x)/(x(1 + x)) - x/(x(1 + x))`

Reducing duplicate factors yields:

`(1 + x - x)/(x(1 + x)) = 1/x - 1/(1 + x)`

Replacing `1/x - 1/(1 + x)` for `1/(x(1 + x))` yields:

`(1 + x - x)/(x^2(1 + x)) = 1/x^2 - 1/x + 1/(1 + x)`

Replacing `1/x^2 - 1/x + 1/(1 + x)` for `1/(x^2(1 + x))` yields:

`(1 + x - x)/(x^3(1 + x)) = 1/x^3 - 1/x^2 + 1/x - 1/(1 + x)`

`1/(x^3(1 + x)) = 1/x^3 - 1/x^2 + 1/x - 1/(1 + x)`

Integrating both sides, yields:

`int 1/(x^3(1 + x)) dx = int 1/x^3 dx - int 1/x^2 dx + int 1/x dx - int 1/(1 + x) dx`

`int 1/(x^3(1 + x)) dx = (x^(-3+1))/(-3+1) - (x^(-2+1))/(-2+1) + ln|x| - ln|1+x| + c`

`int 1/(x^3(1 + x)) dx = -1/(2x^2) + 1/x + ln|x/(x+1)| + c`

**Hence, evaluating the anti-derivative of the given function, yields **`int 1/(x^3+x^4) dx = -1/(2x^2) + 1/x + ln|x/(x+1)| + c.`