What is the integral of (7^x)/(sqrt(1-7^{2x))} ?

The integral of (7^x)/(sqrt(1-7^{2x))} is (arcsin(7^x))/(ln(7))+C

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integral of (7^x)/(sqrt(1-7^{2x))}

\int \frac{7 ^{x}}{1 - 7 ^{2 x}} d x

\frac{arcsin ( 7 ^{x} )}{ln ( 7 )} + C

Solution steps

\int \frac{7 ^{x}}{1 - 7 ^{2 x}} d x

7^{2 x} = 4 9^{x}

= \int \frac{7 ^{x}}{1 - 4 9 ^{x}} d x

Apply u-substitution

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

Take the constant out:

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

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

Apply u-substitution

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = arcsin (v)

= - \frac{1}{2 ln ( 7 )} (- 2 arcsin (v))

Substitute back

= - \frac{1}{2 ln ( 7 )} (- 2 arcsin (7^{\frac{l n ( - ( 1 - 4 9 ^{x} ) + 1 )}{2 l n ( 7 )}}))

Simplify

- \frac{1}{2 ln ( 7 )} (- 2 arcsin (7^{\frac{l n ( - ( 1 - 4 9 ^{x} ) + 1 )}{2 l n ( 7 )}})) : \frac{arcsin ( 7 ^{x} )}{ln ( 7 )}

= \frac{arcsin ( 7 ^{x} )}{ln ( 7 )}

Add a constant to the solution

= \frac{arcsin ( 7 ^{x} )}{ln ( 7 )} + C

\frac{d}{d x} (e^{- 3})

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

\frac{\partial}{\partial x} (cos (x) sin (y))

y^{^{''}} - 10 y^{^{'}} + 61 y = 0

in v erse l a pl a ce \frac{56 s}{s ^{2} + 196}

What is the integral of (7^x)/(sqrt(1-7^{2x))} ?
The integral of (7^x)/(sqrt(1-7^{2x))} is (arcsin(7^x))/(ln(7))+C