What is the integral from 0 to 4 of 1/(x^2-16) ?

The integral from 0 to 4 of 1/(x^2-16) is diverges

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integral from 0 to 4 of 1/(x^2-16)

\int_{0}^{4} \frac{1}{x ^{2} - 16} d x

diverges

Solution steps

\int_{0}^{4} \frac{1}{x ^{2} - 16} d x

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{4 ( u ^{2} - 1 )} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int_{0}^{1} \frac{1}{u ^{2} - 1} d u

Factor

u^{2} - 1 : - (- u^{2} + 1)

= \frac{1}{4} \cdot \int_{0}^{1} \frac{1}{- ( - u ^{2} + 1 )} d u

Take the constant out:

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

= \frac{1}{4} (- \int_{0}^{1} \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}

= \frac{1}{4} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{1})

Simplify

\frac{1}{4} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{1}) : - \frac{1}{4} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{0}^{1}

= - \frac{1}{4} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{0}^{1}

Compute the boundaries:

diverges

= diverges

x \to 4 lim (\frac{5 - x}{x - 4})

6 x, 8 x^{2}, (0, 0.75)

\frac{d}{d x} (c e^{5 t} + c e^{5 y (t, x)})

\int_{4.5}^{5} \frac{x}{8} d x

\frac{1}{4} y^{^{''}} + y^{^{'}} + y = x^{2} - 2 x

What is the integral from 0 to 4 of 1/(x^2-16) ?
The integral from 0 to 4 of 1/(x^2-16) is diverges