What is the integral of 1/(-196.2-4v^2) ?

The integral of 1/(-196.2-4v^2) is -0.03569…arctan(0.14278…v)+C

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integral of 1/(-196.2-4v^2)

\int \frac{1}{- 196.2 - 4 v ^{2}} d v

- 0.03569 \dots arctan (0.14278 \dots v) + C

Solution steps

\int \frac{1}{- 196.2 - 4 v ^{2}} d v

Factor

- 196.2 - 4 v^{2} : - (196.2 + 4 v^{2})

= \int \frac{1}{- ( 196.2 + 4 v ^{2} )} d v

Take the constant out:

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

= - \int \frac{1}{196.2 + 4 v ^{2}} d v

Apply Integral Substitution

= - \int \frac{7.00357 \dots}{196.2 + 196.2 u ^{2}} d u

Take the constant out:

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

= - 7.00357 \dots \cdot \int \frac{1}{196.2 + 196.2 u ^{2}} d u

196.2 + 196.2 u^{2} = 196.2 (1 + \frac{196.2 u ^{2}}{196.2})

= - 7.00357 \dots \cdot \int \frac{1}{196.2 ( 1 + \frac{196.2 u ^{2}}{196.2} )} d u

Take the constant out:

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

= - 7.00357 \dots \cdot \frac{1}{196.2} \cdot \int \frac{1}{1 + \frac{196.2 u ^{2}}{196.2}} d u

Simplify

= - 7.00357 \dots \cdot 0.00509 \dots \cdot \int \frac{1}{1 + \frac{196.2 u ^{2}}{196.2}} d u

Cancel the common factor:

196.2

= - 7.00357 \dots \cdot 0.00509 \dots \cdot \int \frac{1}{1 + u ^{2}} d u

Use the common integral:

\int \frac{1}{1 + u ^{2}} d u = arctan (u)

= - 7.00357 \dots \cdot 0.00509 \dots arctan (u)

Substitute back

u = \frac{2}{14.00714 \dots} v

= - 7.00357 \dots \cdot 0.00509 \dots arctan (\frac{2}{14.00714 \dots} v)

Simplify

= - 0.03569 \dots arctan (0.14278 \dots v)

Add a constant to the solution

= - 0.03569 \dots arctan (0.14278 \dots v) + C

x \to \infty lim (\frac{4}{x})

\frac{d}{d x} ((sec (x))^{3})

\frac{d}{d x} (d cos (t) + t^{2} sin (t))

\frac{d x}{d t} = x^{2} + \frac{1}{25}

d er i v a t i v e f (x) = (x - 1) e^{x}

What is the integral of 1/(-196.2-4v^2) ?
The integral of 1/(-196.2-4v^2) is -0.03569…arctan(0.14278…v)+C