What is the integral from 0 to ln(4) of (e^x)/(sqrt(e^{2x)+25)} ?

The integral from 0 to ln(4) of (e^x)/(sqrt(e^{2x)+25)} is ln(-((4+sqrt(41))(1-sqrt(26)))/(25))

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integral from 0 to ln(4) of (e^x)/(sqrt(e^{2x)+25)}

\int_{0}^{l n (4)} \frac{e ^{x}}{e ^{2 x} + 25} d x

ln (- \frac{( 4 + 41 ) ( 1 - 26 )}{25})

Decimal Notation

0.53397 \dots

Solution steps

\int_{0}^{l n (4)} \frac{e ^{x}}{e ^{2 x} + 25} d x

Apply u-substitution

: \int_{1}^{4} \frac{1}{u ^{2} + 25} d u

= \int_{1}^{4} \frac{1}{u ^{2} + 25} d u

Apply Trigonometric Substitution

: \int_{a r c t a n (\frac{1}{5})}^{a r c t a n (\frac{4}{5})} sec (v) d v

= \int_{a r c t a n (\frac{1}{5})}^{a r c t a n (\frac{4}{5})} 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}{5})}^{a r c t a n (\frac{4}{5})}

Compute the boundaries:

ln (\frac{4}{5} + \frac{41}{5}) - ln (\frac{1}{5} + \frac{26}{5})

= ln (\frac{4}{5} + \frac{41}{5}) - ln (\frac{1}{5} + \frac{26}{5})

Simplify

= ln (- \frac{( 4 + 41 ) ( 1 - 26 )}{25})

\frac{d}{d x} (ln (csc (x) - tan (x)))

x \to - \infty lim (\frac{1}{x} + 3)

\frac{d}{d x} (4 x e^{3 x})

\frac{d}{d x} (\frac{x}{sin ( x ) + cos ( x )})

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

What is the integral from 0 to ln(4) of (e^x)/(sqrt(e^{2x)+25)} ?
The integral from 0 to ln(4) of (e^x)/(sqrt(e^{2x)+25)} is ln(-((4+sqrt(41))(1-sqrt(26)))/(25))