What is the integral from 0 to ln(4) of (e^x)/(sqrt(e^{2x)+81)} ?

The integral from 0 to ln(4) of (e^x)/(sqrt(e^{2x)+81)} is ln(-((4+sqrt(97))(1-sqrt(82)))/(81))

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integral from 0 to ln(4) of (e^x)/(sqrt(e^{2x)+81)}

\int_{0}^{l n (4)} \frac{e ^{x}}{e ^{2 x} + 81} d x

ln (- \frac{( 4 + 97 ) ( 1 - 82 )}{81})

Decimal Notation

0.32009 \dots

Solution steps

\int_{0}^{l n (4)} \frac{e ^{x}}{e ^{2 x} + 81} d x

Apply u-substitution

: \int_{1}^{4} \frac{1}{u ^{2} + 81} d u

= \int_{1}^{4} \frac{1}{u ^{2} + 81} d u

Apply Trigonometric Substitution

: \int_{a r c t a n (\frac{1}{9})}^{a r c t a n (\frac{4}{9})} sec (v) d v

= \int_{a r c t a n (\frac{1}{9})}^{a r c t a n (\frac{4}{9})} sec (v) d v

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= [ln ∣ tan (v) + sec (v) ∣]_{a r c t a n (\frac{1}{9})}^{a r c t a n (\frac{4}{9})}

Compute the boundaries:

ln (\frac{4}{9} + \frac{97}{9}) - ln (\frac{1}{9} + \frac{82}{9})

= ln (\frac{4}{9} + \frac{97}{9}) - ln (\frac{1}{9} + \frac{82}{9})

Simplify

= ln (- \frac{( 4 + 97 ) ( 1 - 82 )}{81})

f (x) = \frac{x ^{3} - 6 x ^{2} + 3}{x ^{2}}

\int tan^{3} (θ) sec^{3} (θ) d θ

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

\int_{- 1}^{1} x^{2} - x - 1 d x

(\frac{4}{3})^{^{'}}

What is the integral from 0 to ln(4) of (e^x)/(sqrt(e^{2x)+81)} ?
The integral from 0 to ln(4) of (e^x)/(sqrt(e^{2x)+81)} is ln(-((4+sqrt(97))(1-sqrt(82)))/(81))