What is the integral of-7/(64+36t^2) ?

The integral of-7/(64+36t^2) is -7/48 arctan((3t)/4)+C

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integral of-7/(64+36t^2)

\int - \frac{7}{64 + 36 t ^{2}} d t

- \frac{7}{48} arctan (\frac{3 t}{4}) + C

Solution steps

\int - \frac{7}{64 + 36 t ^{2}} d t

Take the constant out:

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

= - 7 \cdot \int \frac{1}{64 + 36 t ^{2}} d t

Apply Integral Substitution

= - 7 \cdot \int \frac{1}{48 ( u ^{2} + 1 )} d u

Take the constant out:

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

= - 7 \cdot \frac{1}{48} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= - 7 \cdot \frac{1}{48} arctan (u)

Substitute back

u = \frac{6}{8} t

= - 7 \cdot \frac{1}{48} arctan (\frac{6}{8} t)

Simplify

- 7 \cdot \frac{1}{48} arctan (\frac{6}{8} t) : - \frac{7}{48} arctan (\frac{3 t}{4})

= - \frac{7}{48} arctan (\frac{3 t}{4})

Add a constant to the solution

= - \frac{7}{48} arctan (\frac{3 t}{4}) + C

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

x \to 0 + lim (ln (\frac{2}{x}))

x \to 0 lim ((\frac{1}{x ^{2}})^{2})

\int \frac{x ^{3}}{9 + x ^{8}} d x

k = 1 \sum \infty 0. 5^{k} t

What is the integral of-7/(64+36t^2) ?
The integral of-7/(64+36t^2) is -7/48 arctan((3t)/4)+C