What is the integral of (80)/(64-x^2) ?

The integral of (80)/(64-x^2) is 10(1/2 ln| x/8+1|-1/2 ln| x/8-1|)+C

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integral of (80)/(64-x^2)

\int \frac{80}{64 - x ^{2}} d x

10 (\frac{1}{2} ln \frac{x}{8} + 1 - \frac{1}{2} ln \frac{x}{8} - 1) + C

Solution steps

\int \frac{80}{64 - x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 80 \cdot \int \frac{1}{64 - x ^{2}} d x

Apply Integral Substitution

= 80 \cdot \int \frac{1}{8 ( - u ^{2} + 1 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 80 \cdot \frac{1}{8} \cdot \int \frac{1}{- u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= 80 \cdot \frac{1}{8} (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

Substitute back

u = \frac{x}{8}

= 80 \cdot \frac{1}{8} (\frac{ln \frac{x}{8} + 1}{2} - \frac{ln \frac{x}{8} - 1}{2})

Simplify

80 \cdot \frac{1}{8} (\frac{ln \frac{x}{8} + 1}{2} - \frac{ln \frac{x}{8} - 1}{2}) : 10 (\frac{1}{2} ln \frac{x}{8} + 1 - \frac{1}{2} ln \frac{x}{8} - 1)

= 10 (\frac{1}{2} ln \frac{x}{8} + 1 - \frac{1}{2} ln \frac{x}{8} - 1)

Add a constant to the solution

= 10 (\frac{1}{2} ln \frac{x}{8} + 1 - \frac{1}{2} ln \frac{x}{8} - 1) + C

\frac{\partial}{\partial x} (f (x, b, y) (a x + b y))

5 x, \frac{1}{7} x, 50 - x^{2}

(x^{2} + 1) y^{^{'}} + 7 x (y - 1) = 0, y (0) = 4

x \to \infty lim (cos (0))

sin (x) + \frac{5}{8} cot (x)

What is the integral of (80)/(64-x^2) ?
The integral of (80)/(64-x^2) is 10(1/2 ln| x/8+1|-1/2 ln| x/8-1|)+C