What is the integral of 4xe^{x^2+7}tan(e^{x^2+7}) ?

The integral of 4xe^{x^2+7}tan(e^{x^2+7}) is -2ln|cos(e^{x^2+7})|+C

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integral of 4xe^{x^2+7}tan(e^{x^2+7})

\int 4 x e^{x^{2} + 7} tan (e^{x^{2} + 7}) d x

- 2 ln cos (e^{x^{2} + 7}) + C

Solution steps

\int 4 x e^{x^{2} + 7} tan (e^{x^{2} + 7}) d x

Take the constant out:

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

= 4 \cdot \int x e^{x^{2} + 7} tan (e^{x^{2} + 7}) d x

Apply u-substitution

: \int \frac{e ^{u} tan ( e ^{u} )}{2} d u

= 4 \cdot \int \frac{e ^{u} tan ( e ^{u} )}{2} d u

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \int e^{u} tan (e^{u}) d u

Apply u-substitution

: \int tan (v) d v

= 4 \cdot \frac{1}{2} \cdot \int tan (v) d v

Rewrite using trig identities

= 4 \cdot \frac{1}{2} \cdot \int \frac{sin ( v )}{cos ( v )} d v

Apply u-substitution

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

= 4 \cdot \frac{1}{2} \cdot \int - \frac{1}{w} d w

Take the constant out:

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

= 4 \cdot \frac{1}{2} (- \int \frac{1}{w} d w)

Use the common integral:

\int \frac{1}{w} d w = ln (∣ w ∣)

= 4 \cdot \frac{1}{2} (- ln ∣ w ∣)

Substitute back

= 4 \cdot \frac{1}{2} (- ln cos (e^{x^{2} + 7}))

Simplify

4 \cdot \frac{1}{2} (- ln cos (e^{x^{2} + 7})) : - 2 ln cos (e^{x^{2} + 7})

= - 2 ln cos (e^{x^{2} + 7})

Add a constant to the solution

= - 2 ln cos (e^{x^{2} + 7}) + C

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What is the integral of 4xe^{x^2+7}tan(e^{x^2+7}) ?
The integral of 4xe^{x^2+7}tan(e^{x^2+7}) is -2ln|cos(e^{x^2+7})|+C