What is the integral of 1/(t^2sqrt(256-t^2)) ?

The integral of 1/(t^2sqrt(256-t^2)) is -(sqrt(256-t^2))/(256t)+C

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integral of 1/(t^2sqrt(256-t^2))

\int \frac{1}{t ^{2} 256 - t ^{2}} d t

- \frac{256 - t ^{2}}{256 t} + C

Solution steps

\int \frac{1}{t ^{2} 256 - t ^{2}} d t

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{256 cos ( u ) sin ( u )} d u

= \int \frac{cot ( u )}{256 cos ( u ) sin ( u )} d u

Take the constant out:

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

= \frac{1}{256} \cdot \int \frac{cot ( u )}{cos ( u ) sin ( u )} d u

Simplify

\frac{cot ( u )}{cos ( u ) sin ( u )} : csc^{2} (u)

= \frac{1}{256} \cdot \int csc^{2} (u) d u

Use the common integral:

\int csc^{2} (u) d u = - cot (u)

= \frac{1}{256} (- cot (u))

Substitute back

u = arcsin (\frac{1}{16} t)

= \frac{1}{256} (- cot (arcsin (\frac{1}{16} t)))

Simplify

\frac{1}{256} (- cot (arcsin (\frac{1}{16} t))) : - \frac{256 - t ^{2}}{256 t}

= - \frac{256 - t ^{2}}{256 t}

Add a constant to the solution

= - \frac{256 - t ^{2}}{256 t} + C

n = 0 \sum \infty (\frac{n x}{6})^{n}

x \to + 2 lim (2 - x)

f (x) = (2 t - 1) (3 t - 4)^{- 1}

2^{4 x} + 4 x^{2}

\int \frac{1}{4 ^{2 x}} d x

What is the integral of 1/(t^2sqrt(256-t^2)) ?
The integral of 1/(t^2sqrt(256-t^2)) is -(sqrt(256-t^2))/(256t)+C