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

= \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

: \int - \frac{2}{- v ^{2} + 1} d v

= - \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

cos (a^{7} + x^{7})

\int e^{- 3} d x

x \to \frac{π}{2} lim (cos (2 x))

\int_{0}^{10} x 10 - x d x

y = (x^{2} + 1) (x^{2} - 2 x)

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