What is the integral from 0 to x of 1/(t^2-25) ?

The integral from 0 to x of 1/(t^2-25) is -(ln| x/5+1|-ln| x/5-1|)/(10)

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integral from 0 to x of 1/(t^2-25)

\int_{0}^{x} \frac{1}{t ^{2} - 25} d t

- \frac{ln \frac{x}{5} + 1 - ln \frac{x}{5} - 1}{10}

Solution steps

\int_{0}^{x} \frac{1}{t ^{2} - 25} d t

Apply Integral Substitution

= \int_{0}^{\frac{x}{5}} \frac{1}{5 ( 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}{5} \cdot \int_{0}^{\frac{x}{5}} \frac{1}{u ^{2} - 1} d u

Factor

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

= \frac{1}{5} \cdot \int_{0}^{\frac{x}{5}} \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}{5} (- \int_{0}^{\frac{x}{5}} \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}{5} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{\frac{x}{5}})

Simplify

\frac{1}{5} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{\frac{x}{5}}) : - \frac{1}{5} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{0}^{\frac{x}{5}}

= - \frac{1}{5} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{0}^{\frac{x}{5}}

Compute the boundaries:

\frac{ln \frac{x}{5} + 1 - ln \frac{x}{5} - 1}{2}

= - \frac{1}{5} \cdot \frac{ln \frac{x}{5} + 1 - ln \frac{x}{5} - 1}{2}

Simplify

= - \frac{ln \frac{x}{5} + 1 - ln \frac{x}{5} - 1}{10}

y^{^{'}} = y + 3 t e^{t}, y (0) = 4

x \to 3 lim (x - 8)

\int (s^{3} + 3)^{2} (1 - s^{2})^{3} d s

y^{^{''}} - 16 y = 2 e^{4 x}

\int_{1}^{2} t ln (t) d t

What is the integral from 0 to x of 1/(t^2-25) ?
The integral from 0 to x of 1/(t^2-25) is -(ln| x/5+1|-ln| x/5-1|)/(10)