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

The integral of ((x-4)^2(x+1))/x is (x^3)/3-(7x^2)/2+8x+16ln|x|+C

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integral of ((x-4)^2(x+1))/x

\int \frac{( x - 4 ) ^{2} ( x + 1 )}{x} d x

\frac{x ^{3}}{3} - \frac{7 x ^{2}}{2} + 8 x + 16 ln ∣ x ∣ + C

Solution steps

\int \frac{( x - 4 ) ^{2} ( x + 1 )}{x} d x

Expand

\frac{( x - 4 ) ^{2} ( x + 1 )}{x} : x^{2} - 7 x + 8 + \frac{16}{x}

= \int x^{2} - 7 x + 8 + \frac{16}{x} 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^{2} d x - \int 7 x d x + \int 8 d x + \int \frac{16}{x} d x

\int x^{2} d x = \frac{x ^{3}}{3}

\int 7 x d x = \frac{7 x ^{2}}{2}

\int 8 d x = 8 x

\int \frac{16}{x} d x = 16 ln ∣ x ∣

= \frac{x ^{3}}{3} - \frac{7 x ^{2}}{2} + 8 x + 16 ln ∣ x ∣

Add a constant to the solution

= \frac{x ^{3}}{3} - \frac{7 x ^{2}}{2} + 8 x + 16 ln ∣ x ∣ + C

x y^{^{'}} = 2 y + 8 x

\int_{2}^{3} \frac{x ^{2} + 1}{x ^{3} + 3 x} d x

\int_{0}^{5} \frac{x}{x - 2} d x

\int 3 x sec^{2} (4) x^{2} d x

\frac{d x}{d t} = r x

What is the integral of ((x-4)^2(x+1))/x ?
The integral of ((x-4)^2(x+1))/x is (x^3)/3-(7x^2)/2+8x+16ln|x|+C