What is the integral of sqrt(4t^2+5) ?

The integral of sqrt(4t^2+5) is 1/4 (2tsqrt(4t^2+5)+5(ln|2t+sqrt(5)sqrt((4t^2+5)/5)|-(ln(5))/2))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sqrt(4t^2+5)

\int 4 t^{2} + 5 d t

\frac{1}{4} (2 t 4 t^{2} + 5 + 5 (ln 2 t + 5 \frac{4 t ^{2} + 5}{5} - \frac{ln ( 5 )}{2})) + C

Solution steps

\int 4 t^{2} + 5 d t

Apply Trigonometric Substitution

: \int \frac{5}{2} sec^{3} (u) d u

= \int \frac{5}{2} sec^{3} (u) d u

Take the constant out:

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

= \frac{5}{2} \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= \frac{5}{2} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} \cdot \int sec (u) d u)

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

= \frac{5}{2} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (\frac{2}{5} t)

= \frac{5}{2} \frac{sec ^{2} ( arctan ( \frac{2}{5} t ) ) sin ( arctan ( \frac{2}{5} t ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{2}{5} t)) + sec (arctan (\frac{2}{5} t))

Simplify

\frac{5}{2} \frac{sec ^{2} ( arctan ( \frac{2}{5} t ) ) sin ( arctan ( \frac{2}{5} t ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{2}{5} t)) + sec (arctan (\frac{2}{5} t)) : \frac{1}{4} (2 t 4 t^{2} + 5 + 5 (ln 2 t + 5 \frac{4 t ^{2} + 5}{5} - \frac{ln ( 5 )}{2}))

= \frac{1}{4} (2 t 4 t^{2} + 5 + 5 (ln 2 t + 5 \frac{4 t ^{2} + 5}{5} - \frac{ln ( 5 )}{2}))

Add a constant to the solution

= \frac{1}{4} (2 t 4 t^{2} + 5 + 5 (ln 2 t + 5 \frac{4 t ^{2} + 5}{5} - \frac{ln ( 5 )}{2})) + C

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

\int \frac{3}{x} + sin (2 x) d x

\int x^{2} - 8 d x

\int x^{\frac{5}{2}} - x^{\frac{1}{2}} d x

\int x^{2} (2 x^{3} + 4) d x

What is the integral of sqrt(4t^2+5) ?
The integral of sqrt(4t^2+5) is 1/4 (2tsqrt(4t^2+5)+5(ln|2t+sqrt(5)sqrt((4t^2+5)/5)|-(ln(5))/2))+C