What is the integral of (e^t)/(sqrt(e^{2t)+9)} ?

The integral of (e^t)/(sqrt(e^{2t)+9)} is ln(1/3 |e^t+sqrt(9+e^{2t)}|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (e^t)/(sqrt(e^{2t)+9)}

\int \frac{e ^{t}}{e ^{2 t} + 9} d t

ln (\frac{1}{3} e^{t} + 9 + e^{2 t}) + C

Solution steps

\int \frac{e ^{t}}{e ^{2 t} + 9} d t

Apply u-substitution

: \int \frac{1}{u ^{2} + 9} d u

= \int \frac{1}{u ^{2} + 9} d u

Apply Trigonometric Substitution

: \int sec (v) d v

= \int sec (v) d v

Use the common integral:

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

= ln ∣ tan (v) + sec (v) ∣

Substitute back

= ln tan (arctan (\frac{1}{3} e^{t})) + sec (arctan (\frac{1}{3} e^{t}))

Simplify

ln tan (arctan (\frac{1}{3} e^{t})) + sec (arctan (\frac{1}{3} e^{t})) : ln (\frac{1}{3} e^{t} + 9 + e^{2 t})

= ln (\frac{1}{3} e^{t} + 9 + e^{2 t})

Add a constant to the solution

= ln (\frac{1}{3} e^{t} + 9 + e^{2 t}) + C

\int 4 x^{2} arctan (x) d x

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

\int 10 x^{\frac{2}{3}} d x

\int (5 x - 2) sin (x) d x

\int \frac{cos ^{3} ( w )}{cos ^{4} ( w )} d w

What is the integral of (e^t)/(sqrt(e^{2t)+9)} ?
The integral of (e^t)/(sqrt(e^{2t)+9)} is ln(1/3 |e^t+sqrt(9+e^{2t)}|)+C