What is the integral of sec^2(3x+1) ?

The integral of sec^2(3x+1) is 1/3 tan(3x+1)+C

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integral of sec^2(3x+1)

\int sec^{2} (3 x + 1) d x

\frac{1}{3} tan (3 x + 1) + C

Solution steps

\int sec^{2} (3 x + 1) d x

Apply u-substitution

= \int sec^{2} (u) \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int sec^{2} (u) d u

Use the common integral:

\int sec^{2} (u) d u = tan (u)

= \frac{1}{3} tan (u)

Substitute back

u = 3 x + 1

= \frac{1}{3} tan (3 x + 1)

Add a constant to the solution

= \frac{1}{3} tan (3 x + 1) + C

What is the integral of sec^2(3x+1) ?
The integral of sec^2(3x+1) is 1/3 tan(3x+1)+C