What is the integral of sqrt(9+t^2) ?

The integral of sqrt(9+t^2) is 9(1/18 tsqrt(9+t^2)+1/2 ln((|t+sqrt(9+t^2)|)/3))+C

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integral of sqrt(9+t^2)

\int 9 + t^{2} d t

9 (\frac{1}{18} t 9 + t^{2} + \frac{1}{2} ln (\frac{t + 9 + t ^{2}}{3})) + C

Solution steps

\int 9 + t^{2} d t

Apply Trigonometric Substitution

: \int 9 sec^{3} (u) d u

= \int 9 sec^{3} (u) d u

Take the constant out:

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

= 9 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 9 (\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) ∣

= 9 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (\frac{1}{3} t)

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

Simplify

9 (\frac{sec ^{2} ( arctan ( \frac{1}{3} t ) ) sin ( arctan ( \frac{1}{3} t ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{3} t)) + sec (arctan (\frac{1}{3} t))) : 9 (\frac{1}{18} t 9 + t^{2} + \frac{1}{2} ln (\frac{t + 9 + t ^{2}}{3}))

= 9 (\frac{1}{18} t 9 + t^{2} + \frac{1}{2} ln (\frac{t + 9 + t ^{2}}{3}))

Add a constant to the solution

= 9 (\frac{1}{18} t 9 + t^{2} + \frac{1}{2} ln (\frac{t + 9 + t ^{2}}{3})) + C

\frac{d y}{d x} = \frac{y}{x} + \frac{9 x ^{2}}{y ^{2}}

\int \frac{12 x ^{2}}{4 x ^{3} + 1} d x

\int \frac{2 x - 5}{x ^{2} - 5 x + 6} d x

\int_{12}^{\infty} x e^{- x} d x

f (x) = sin (e^{x})

What is the integral of sqrt(9+t^2) ?
The integral of sqrt(9+t^2) is 9(1/18 tsqrt(9+t^2)+1/2 ln((|t+sqrt(9+t^2)|)/3))+C