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

The integral of x-1-(x-1)^4 is (x^2)/2-x-((x-1)^5)/5+C

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integral of x-1-(x-1)^4

\int x - 1 - (x - 1)^{4} d x

\frac{x ^{2}}{2} - x - \frac{( x - 1 ) ^{5}}{5} + C

Solution steps

\int x - 1 - (x - 1)^{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 x d x - \int 1 d x - \int (x - 1)^{4} d x

\int x d x = \frac{x ^{2}}{2}

\int 1 d x = x

\int (x - 1)^{4} d x = \frac{( x - 1 ) ^{5}}{5}

= \frac{x ^{2}}{2} - x - \frac{( x - 1 ) ^{5}}{5}

Add a constant to the solution

= \frac{x ^{2}}{2} - x - \frac{( x - 1 ) ^{5}}{5} + C

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What is the integral of x-1-(x-1)^4 ?
The integral of x-1-(x-1)^4 is (x^2)/2-x-((x-1)^5)/5+C