What is the integral of 1/(tsqrt(2t^2-9)) ?

The integral of 1/(tsqrt(2t^2-9)) is 1/3 arctan(1/3 sqrt(2t^2-9))+C

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integral of 1/(tsqrt(2t^2-9))

\int \frac{1}{t 2 t ^{2} - 9} d t

\frac{1}{3} arctan (\frac{1}{3} 2 t^{2} - 9) + C

Solution steps

\int \frac{1}{t 2 t ^{2} - 9} d t

Apply u-substitution

: \int \frac{1}{2 u 2 u - 9} d u

= \int \frac{1}{2 u 2 u - 9} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u 2 u - 9} d u

Apply u-substitution

: \int \frac{2}{v ^{2} + 9} d v

= \frac{1}{2} \cdot \int \frac{2}{v ^{2} + 9} d v

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \int \frac{1}{v ^{2} + 9} d v

Apply Integral Substitution

= \frac{1}{2} \cdot 2 \cdot \int \frac{1}{3 ( w ^{2} + 1 )} d w

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \frac{1}{3} \cdot \int \frac{1}{w ^{2} + 1} d w

Use the common integral:

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

= \frac{1}{2} \cdot 2 \cdot \frac{1}{3} arctan (w)

Substitute back

= \frac{1}{2} \cdot 2 \cdot \frac{1}{3} arctan (\frac{2 t ^{2} - 9}{3})

Simplify

\frac{1}{2} \cdot 2 \cdot \frac{1}{3} arctan (\frac{2 t ^{2} - 9}{3}) : \frac{1}{3} arctan (\frac{1}{3} 2 t^{2} - 9)

= \frac{1}{3} arctan (\frac{1}{3} 2 t^{2} - 9)

Add a constant to the solution

= \frac{1}{3} arctan (\frac{1}{3} 2 t^{2} - 9) + C

x \to 0 lim (\frac{( tan ( 3 x ) )}{x})

\frac{\partial}{\partial y} (sin (x^{2} - y))

\int (tan^{3} (x)) d x

y = x^{2}, y = x + 2, 0

y = \frac{x}{x ^{2} - 4 x + 5}, y = \frac{3}{x ^{2} - 4 x + 5}, x = 0

What is the integral of 1/(tsqrt(2t^2-9)) ?
The integral of 1/(tsqrt(2t^2-9)) is 1/3 arctan(1/3 sqrt(2t^2-9))+C