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

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

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integral of sec(2x+1)tan(2x+1)

\int sec (2 x + 1) tan (2 x + 1) d x

\frac{1}{2} sec (2 x + 1) + C

Solution steps

\int sec (2 x + 1) tan (2 x + 1) d x

Apply u-substitution

: \int sec (u) tan (u) \frac{1}{2} d u

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

Take the constant out:

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

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

Apply u-substitution

: \int 1 d v

= \frac{1}{2} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot 1 \cdot v

Substitute back

= \frac{1}{2} \cdot 1 \cdot sec (2 x + 1)

Simplify

= \frac{1}{2} sec (2 x + 1)

Add a constant to the solution

= \frac{1}{2} sec (2 x + 1) + C

x \to 1 + lim (ln^{x - 1} (x))

\frac{d}{d t} (- cos^{2} (t))

\int \frac{x ^{9}}{1 + x ^{20}} d x

4 e^{- 3 t + 2}

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