What is the integral of 1/(sqrt(16e^{6x)-25)} ?

The integral of 1/(sqrt(16e^{6x)-25)} is 1/15 arctan(1/5 sqrt(16e^{6x)-25})+C

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integral of 1/(sqrt(16e^{6x)-25)}

\int \frac{1}{16 e ^{6 x} - 25} d x

\frac{1}{15} arctan (\frac{1}{5} 16 e^{6 x} - 25) + C

Solution steps

\int \frac{1}{16 e ^{6 x} - 25} d x

Apply u-substitution

= \int \frac{1}{6 u ( u + 25 )} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{1}{u ( u + 25 )} d u

Apply u-substitution

= \frac{1}{6} \cdot \int \frac{2}{v ^{2} + 25} d v

Take the constant out:

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

= \frac{1}{6} \cdot 2 \cdot \int \frac{1}{v ^{2} + 25} d v

Apply Integral Substitution

= \frac{1}{6} \cdot 2 \cdot \int \frac{1}{5 ( 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}{6} \cdot 2 \cdot \frac{1}{5} \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}{6} \cdot 2 \cdot \frac{1}{5} arctan (w)

Substitute back

= \frac{1}{6} \cdot 2 \cdot \frac{1}{5} arctan (\frac{16 e ^{6 x} - 25}{5})

Simplify

\frac{1}{6} \cdot 2 \cdot \frac{1}{5} arctan (\frac{16 e ^{6 x} - 25}{5}) : \frac{1}{15} arctan (\frac{1}{5} 16 e^{6 x} - 25)

= \frac{1}{15} arctan (\frac{1}{5} 16 e^{6 x} - 25)

Add a constant to the solution

= \frac{1}{15} arctan (\frac{1}{5} 16 e^{6 x} - 25) + C

What is the integral of 1/(sqrt(16e^{6x)-25)} ?
The integral of 1/(sqrt(16e^{6x)-25)} is 1/15 arctan(1/5 sqrt(16e^{6x)-25})+C