What is the integral of (2x-3)tan(7x^2-21x+9) ?

The integral of (2x-3)tan(7x^2-21x+9) is -1/7 ln|cos(7x^2-21x+9)|+C

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integral of (2x-3)tan(7x^2-21x+9)

\int (2 x - 3) tan (7 x^{2} - 21 x + 9) d x

- \frac{1}{7} ln cos (7 x^{2} - 21 x + 9) + C

Solution steps

\int (2 x - 3) tan (7 x^{2} - 21 x + 9) d x

Apply u-substitution

: \int \frac{tan ( u )}{7} d u

= \int \frac{tan ( u )}{7} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int tan (u) d u

Rewrite using trig identities

= \frac{1}{7} \cdot \int \frac{sin ( u )}{cos ( u )} d u

Apply u-substitution

: \int - \frac{1}{v} d v

= \frac{1}{7} \cdot \int - \frac{1}{v} d v

Take the constant out:

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

= \frac{1}{7} (- \int \frac{1}{v} d v)

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{7} (- ln ∣ v ∣)

Substitute back

= \frac{1}{7} (- ln cos (7 x^{2} - 21 x + 9))

Simplify

= - \frac{1}{7} ln cos (7 x^{2} - 21 x + 9)

Add a constant to the solution

= - \frac{1}{7} ln cos (7 x^{2} - 21 x + 9) + C

y = \frac{x + 6}{x ^{3} + x - 7}

(y^{4} - x^{2}) \frac{d y}{d x} = - x \cdot y

(x^{2} + 1) (\frac{d y}{d x}) + 3 x (y - 1) = 0

\frac{\partial}{\partial x} (7 x ln (x y))

\frac{d}{d x} (1 3^{x})

What is the integral of (2x-3)tan(7x^2-21x+9) ?
The integral of (2x-3)tan(7x^2-21x+9) is -1/7 ln|cos(7x^2-21x+9)|+C