What is the integral of 1/(sqrt(4^x+1)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(sqrt(4^x+1))

\int \frac{1}{4 ^{x} + 1} d x

- \frac{ln 4 ^{x} + 1 + 1 - ln 4 ^{x} + 1 - 1}{2 ln ( 2 )} + C

Solution steps

\int \frac{1}{4 ^{x} + 1} d x

Apply u-substitution

: \int \frac{1}{ln ( 2 ) ( u ^{2} - 1 )} d u

= \int \frac{1}{ln ( 2 ) ( u ^{2} - 1 )} d u

Take the constant out:

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

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

Factor

u^{2} - 1 : - (- u^{2} + 1)

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= \frac{1}{ln ( 2 )} (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

Substitute back

u = 4^{x} + 1

= \frac{1}{ln ( 2 )} (- (\frac{ln 4 ^{x} + 1 + 1}{2} - \frac{ln 4 ^{x} + 1 - 1}{2}))

Simplify

\frac{1}{ln ( 2 )} (- (\frac{ln 4 ^{x} + 1 + 1}{2} - \frac{ln 4 ^{x} + 1 - 1}{2})) : - \frac{ln 4 ^{x} + 1 + 1 - ln 4 ^{x} + 1 - 1}{2 ln ( 2 )}

= - \frac{ln 4 ^{x} + 1 + 1 - ln 4 ^{x} + 1 - 1}{2 ln ( 2 )}

Add a constant to the solution

= - \frac{ln 4 ^{x} + 1 + 1 - ln 4 ^{x} + 1 - 1}{2 ln ( 2 )} + C

\int 9 - r^{2} d r

\int - x^{4} e^{x^{5}} d x

\int (9 - x^{2})^{3} d x

\int cos^{5} (\frac{x}{3}) d x

\int cos (x) \frac{1}{x} d x

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