What is the integral of (48)/(36-x^2) ?

The integral of (48)/(36-x^2) is 8(1/2 ln| x/6+1|-1/2 ln| x/6-1|)+C

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integral of (48)/(36-x^2)

\int \frac{48}{36 - x ^{2}} d x

8 (\frac{1}{2} ln \frac{x}{6} + 1 - \frac{1}{2} ln \frac{x}{6} - 1) + C

Solution steps

\int \frac{48}{36 - x ^{2}} d x

Take the constant out:

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

= 48 \cdot \int \frac{1}{36 - x ^{2}} d x

Apply Integral Substitution

= 48 \cdot \int \frac{1}{6 ( - u ^{2} + 1 )} d u

Take the constant out:

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

= 48 \cdot \frac{1}{6} \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}

= 48 \cdot \frac{1}{6} (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

Substitute back

u = \frac{x}{6}

= 48 \cdot \frac{1}{6} (\frac{ln \frac{x}{6} + 1}{2} - \frac{ln \frac{x}{6} - 1}{2})

Simplify

48 \cdot \frac{1}{6} (\frac{ln \frac{x}{6} + 1}{2} - \frac{ln \frac{x}{6} - 1}{2}) : 8 (\frac{1}{2} ln \frac{x}{6} + 1 - \frac{1}{2} ln \frac{x}{6} - 1)

= 8 (\frac{1}{2} ln \frac{x}{6} + 1 - \frac{1}{2} ln \frac{x}{6} - 1)

Add a constant to the solution

= 8 (\frac{1}{2} ln \frac{x}{6} + 1 - \frac{1}{2} ln \frac{x}{6} - 1) + C

\frac{d}{d x} (\frac{( x + b ) ( x - a )}{( x - b )})

f (x) = 4 x + 84, (4, 10)

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

\int_{0}^{x^{2}} sin (t^{3}) d t

\frac{d}{d x} (cos (a^{3} + x^{3}))

What is the integral of (48)/(36-x^2) ?
The integral of (48)/(36-x^2) is 8(1/2 ln| x/6+1|-1/2 ln| x/6-1|)+C