What is the integral of 1/(sqrt((5t+1)^2-1)) ?

The integral of 1/(sqrt((5t+1)^2-1)) is 1/5 ln|sqrt(25t^2+10t)+5t+1|+C

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integral of 1/(sqrt((5t+1)^2-1))

\int \frac{1}{( 5 t + 1 ) ^{2} - 1} d t

\frac{1}{5} ln 25 t^{2} + 10 t + 5 t + 1 + C

Solution steps

\int \frac{1}{( 5 t + 1 ) ^{2} - 1} d t

Expand

(5 t + 1)^{2} - 1 : 25 t^{2} + 10 t

= \int \frac{1}{25 t ^{2} + 10 t} d t

Complete the square

25 t^{2} + 10 t : 25 (t + \frac{1}{5})^{2} - 1

= \int \frac{1}{25 ( t + \frac{1}{5} ) ^{2} - 1} d t

Apply u-substitution

= \int \frac{1}{25 u ^{2} - 1} d u

Apply Trigonometric Substitution

= \int \frac{sec ( v )}{5} d v

Take the constant out:

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

= \frac{1}{5} \cdot \int sec (v) d v

Use the common integral:

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

= \frac{1}{5} ln ∣ tan (v) + sec (v) ∣

Substitute back

= \frac{1}{5} ln tan (arcsec (5 (t + \frac{1}{5}))) + sec (arcsec (5 (t + \frac{1}{5})))

Simplify

\frac{1}{5} ln tan (arcsec (5 (t + \frac{1}{5}))) + sec (arcsec (5 (t + \frac{1}{5}))) : \frac{1}{5} ln 25 t^{2} + 10 t + 5 t + 1

= \frac{1}{5} ln 25 t^{2} + 10 t + 5 t + 1

Add a constant to the solution

= \frac{1}{5} ln 25 t^{2} + 10 t + 5 t + 1 + C

What is the integral of 1/(sqrt((5t+1)^2-1)) ?
The integral of 1/(sqrt((5t+1)^2-1)) is 1/5 ln|sqrt(25t^2+10t)+5t+1|+C