Solve ∫ (x-1)^{3}+(x-1)^{2}-x+x(4-x)(4+x)+(8-x-x^2)^2+x^2(17-x^2)

$$\int ( x - 1 ) ^ { 3 } + ( x - 1 ) ^ { 2 } - x + x ( 4 - x ) ( 4 + x ) + ( 8 - x - x ^ { 2 } ) ^ { 2 } + x ^ { 2 } ( 17 - x ^ { 2 } )$$

Evaluate

$$\frac{x^{4}}{2}+64x+С$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(x-1\right)^{3}.

$$\int x^{3}-3x^{2}+3x-1+\left(x-1\right)^{2}-x+x\left(4-x\right)\left(4+x\right)+\left(8-x-x^{2}\right)^{2}+x^{2}\left(17-x^{2}\right)\mathrm{d}x$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$\int x^{3}-3x^{2}+3x-1+x^{2}-2x+1-x+x\left(4-x\right)\left(4+x\right)+\left(8-x-x^{2}\right)^{2}+x^{2}\left(17-x^{2}\right)\mathrm{d}x$$

Combine -3x^{2} and x^{2} to get -2x^{2}.

$$\int x^{3}-2x^{2}+3x-1-2x+1-x+x\left(4-x\right)\left(4+x\right)+\left(8-x-x^{2}\right)^{2}+x^{2}\left(17-x^{2}\right)\mathrm{d}x$$

Combine 3x and -2x to get x.

$$\int x^{3}-2x^{2}+x-1+1-x+x\left(4-x\right)\left(4+x\right)+\left(8-x-x^{2}\right)^{2}+x^{2}\left(17-x^{2}\right)\mathrm{d}x$$

Add -1 and 1 to get 0.

$$\int x^{3}-2x^{2}+x-x+x\left(4-x\right)\left(4+x\right)+\left(8-x-x^{2}\right)^{2}+x^{2}\left(17-x^{2}\right)\mathrm{d}x$$

Use the distributive property to multiply x by 4-x.

$$\int x^{3}-2x^{2}+x-x+\left(4x-x^{2}\right)\left(4+x\right)+\left(8-x-x^{2}\right)^{2}+x^{2}\left(17-x^{2}\right)\mathrm{d}x$$

Use the distributive property to multiply 4x-x^{2} by 4+x and combine like terms.

$$\int x^{3}-2x^{2}+x-x+16x-x^{3}+\left(8-x-x^{2}\right)^{2}+x^{2}\left(17-x^{2}\right)\mathrm{d}x$$

Combine x and 16x to get 17x.

$$\int x^{3}-2x^{2}+17x-x-x^{3}+\left(8-x-x^{2}\right)^{2}+x^{2}\left(17-x^{2}\right)\mathrm{d}x$$

Combine x^{3} and -x^{3} to get 0.

$$\int -2x^{2}+17x-x+\left(8-x-x^{2}\right)^{2}+x^{2}\left(17-x^{2}\right)\mathrm{d}x$$

Square 8-x-x^{2}.

$$\int -2x^{2}+17x-x+x^{4}+2x^{3}-15x^{2}-16x+64+x^{2}\left(17-x^{2}\right)\mathrm{d}x$$

Combine -2x^{2} and -15x^{2} to get -17x^{2}.

$$\int -17x^{2}+17x-x+x^{4}+2x^{3}-16x+64+x^{2}\left(17-x^{2}\right)\mathrm{d}x$$

Combine 17x and -16x to get x.

$$\int -17x^{2}+x-x+x^{4}+2x^{3}+64+x^{2}\left(17-x^{2}\right)\mathrm{d}x$$

Use the distributive property to multiply x^{2} by 17-x^{2}.

$$\int -17x^{2}+x-x+x^{4}+2x^{3}+64+17x^{2}-x^{4}\mathrm{d}x$$

Combine -17x^{2} and 17x^{2} to get 0.

$$\int x-x+x^{4}+2x^{3}+64-x^{4}\mathrm{d}x$$

Combine x^{4} and -x^{4} to get 0.

$$\int x-x+2x^{3}+64\mathrm{d}x$$

Combine x and -x to get 0.

$$\int 2x^{3}+64\mathrm{d}x$$

Integrate the sum term by term.

$$\int 2x^{3}\mathrm{d}x+\int 64\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2\int x^{3}\mathrm{d}x+\int 64\mathrm{d}x$$

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}. Multiply 2 times \frac{x^{4}}{4}.

$$\frac{x^{4}}{2}+\int 64\mathrm{d}x$$

Find the integral of 64 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{4}}{2}+64x$$

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.

$$64x+\frac{x^{4}}{2}+С$$

Differentiate w.r.t. x

$$2\left(x^{3}+32\right)$$