Evaluate
\frac{x^{4}}{4}-5x^{3}+37x^{2}-120x+С
Differentiate w.r.t. x
\left(x-6\right)\left(x-5\right)\left(x-4\right)
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\int \left(x^{2}-5x-6x+30\right)\left(x-4\right)\mathrm{d}x
Apply the distributive property by multiplying each term of x-6 by each term of x-5.
\int \left(x^{2}-11x+30\right)\left(x-4\right)\mathrm{d}x
Combine -5x and -6x to get -11x.
\int x^{3}-4x^{2}-11x^{2}+44x+30x-120\mathrm{d}x
Apply the distributive property by multiplying each term of x^{2}-11x+30 by each term of x-4.
\int x^{3}-15x^{2}+44x+30x-120\mathrm{d}x
Combine -4x^{2} and -11x^{2} to get -15x^{2}.
\int x^{3}-15x^{2}+74x-120\mathrm{d}x
Combine 44x and 30x to get 74x.
\int x^{3}\mathrm{d}x+\int -15x^{2}\mathrm{d}x+\int 74x\mathrm{d}x+\int -120\mathrm{d}x
Integrate the sum term by term.
\int x^{3}\mathrm{d}x-15\int x^{2}\mathrm{d}x+74\int x\mathrm{d}x+\int -120\mathrm{d}x
Factor out the constant in each of the terms.
\frac{x^{4}}{4}-15\int x^{2}\mathrm{d}x+74\int x\mathrm{d}x+\int -120\mathrm{d}x
Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}.
\frac{x^{4}}{4}-5x^{3}+74\int x\mathrm{d}x+\int -120\mathrm{d}x
Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply -15 times \frac{x^{3}}{3}.
\frac{x^{4}}{4}-5x^{3}+37x^{2}+\int -120\mathrm{d}x
Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply 74 times \frac{x^{2}}{2}.
\frac{x^{4}}{4}-5x^{3}+37x^{2}-120x
Find the integral of -120 using the table of common integrals rule \int a\mathrm{d}x=ax.
\frac{x^{4}}{4}-5x^{3}+37x^{2}-120x+С
If F\left(x\right) is an antiderivative of f\left(x\right), then the set of all antiderivatives of f\left(x\right) is given by F\left(x\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.
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