What is the integral of (4t)/((2-8t^2)) ?

The integral of (4t)/((2-8t^2)) is -1/4 ln|2-8t^2|+C

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integral of (4t)/((2-8t^2))

integral

\frac{4 t}{( 2 - 8 t ^{2} )}

- \frac{1}{4} ln 2 - 8 t^{2} + C

Solution steps

\int \frac{4 t}{2 - 8 t ^{2}} d t

Take the constant out:

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

= 4 \cdot \int \frac{t}{2 - 8 t ^{2}} d t

Apply u-substitution

: \int - \frac{1}{16 u} d u

= 4 \cdot \int - \frac{1}{16 u} d u

Take the constant out:

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

= 4 (- \frac{1}{16} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 (- \frac{1}{16} ln ∣ u ∣)

Substitute back

u = 2 - 8 t^{2}

= 4 (- \frac{1}{16} ln 2 - 8 t^{2})

Simplify

4 (- \frac{1}{16} ln 2 - 8 t^{2}) : - \frac{1}{4} ln 2 - 8 t^{2}

= - \frac{1}{4} ln 2 - 8 t^{2}

Add a constant to the solution

= - \frac{1}{4} ln 2 - 8 t^{2} + C

\frac{( 1 + cos ^{2} ( x ) )}{cos ^{2} ( x )}

(e^{x})^{\frac{1}{2}}

\frac{1}{( 2 x + 1 ) ^{2}}

cos (2 - 3 x)

\frac{( x - 1 ) ^{2}}{x}

What is the integral of (4t)/((2-8t^2)) ?
The integral of (4t)/((2-8t^2)) is -1/4 ln|2-8t^2|+C