What is the integral of 1/(sqrt(49+t^2)) ?

The integral of 1/(sqrt(49+t^2)) is ln(1/7 |t+sqrt(49+t^2)|)+C

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integral of 1/(sqrt(49+t^2))

\int \frac{1}{49 + t ^{2}} d t

ln (\frac{1}{7} t + 49 + t^{2}) + C

Solution steps

\int \frac{1}{49 + t ^{2}} d t

Apply Trigonometric Substitution

: \int sec (u) d u

= \int sec (u) d u

Use the common integral:

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

= ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arctan (\frac{1}{7} t)

= ln tan (arctan (\frac{1}{7} t)) + sec (arctan (\frac{1}{7} t))

Simplify

ln tan (arctan (\frac{1}{7} t)) + sec (arctan (\frac{1}{7} t)) : ln (\frac{1}{7} t + 49 + t^{2})

= ln (\frac{1}{7} t + 49 + t^{2})

Add a constant to the solution

= ln (\frac{1}{7} t + 49 + t^{2}) + C

y = sin (x), at x = \frac{7 π}{6}

x \to 0.9 lim (2 x^{2} - 5 x + 1)

\int - \frac{x}{x + 1} d x

f (x) = \frac{1}{x ^{6}}, (- 2, \frac{1}{64})

\int_{0}^{1} \frac{4}{x ^{2} + 3 x + 2} d x

What is the integral of 1/(sqrt(49+t^2)) ?
The integral of 1/(sqrt(49+t^2)) is ln(1/7 |t+sqrt(49+t^2)|)+C