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

The integral of (x+1)^4-(x+1)^3 is ((x+1)^5)/5-((x+1)^4)/4+C

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integral of (x+1)^4-(x+1)^3

\int (x + 1)^{4} - (x + 1)^{3} d x

\frac{( x + 1 ) ^{5}}{5} - \frac{( x + 1 ) ^{4}}{4} + C

Solution steps

\int (x + 1)^{4} - (x + 1)^{3} 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 + 1)^{4} d x - \int (x + 1)^{3} d x

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

\int (x + 1)^{3} d x = \frac{( x + 1 ) ^{4}}{4}

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

Add a constant to the solution

= \frac{( x + 1 ) ^{5}}{5} - \frac{( x + 1 ) ^{4}}{4} + C

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What is the integral of (x+1)^4-(x+1)^3 ?
The integral of (x+1)^4-(x+1)^3 is ((x+1)^5)/5-((x+1)^4)/4+C