What is the integral of (x^4+8x^2+8)/(x^3-4x) ?

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

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integral of (x^4+8x^2+8)/(x^3-4x)

\int \frac{x ^{4} + 8 x ^{2} + 8}{x ^{3} - 4 x} d x

\frac{x ^{2}}{2} + 7 ln x^{2} - 4 - 2 ln (x) + C

Solution steps

\int \frac{x ^{4} + 8 x ^{2} + 8}{x ^{3} - 4 x} d x

Long division

\frac{x ^{4} + 8 x ^{2} + 8}{x ^{3} - 4 x} : x + \frac{12 x ^{2} + 8}{x ^{3} - 4 x}

= \int x + \frac{12 x ^{2} + 8}{x ^{3} - 4 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 d x + \int \frac{12 x ^{2} + 8}{x ^{3} - 4 x} d x

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

\int \frac{12 x ^{2} + 8}{x ^{3} - 4 x} d x = 7 ln x^{2} - 4 - 2 ln (x)

= \frac{x ^{2}}{2} + 7 ln x^{2} - 4 - 2 ln (x)

Add a constant to the solution

= \frac{x ^{2}}{2} + 7 ln x^{2} - 4 - 2 ln (x) + C

x^{3} - 3 x^{2} + 5 x - 7, 0

(1.7) (3.3)

\int_{0}^{2} \frac{d t}{4 + t ^{2}} d x

y^{^{''}} + 49 y = 0, y (0) = 4, y^{^{'}} (0) = - 6

y^{^{'}} - y = 3 x - 3

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