What is the integral of (4x+e^x)sin(6x^2+3e^x) ?

The integral of (4x+e^x)sin(6x^2+3e^x) is -1/3 cos(6x^2+3e^x)+C

Popular >

integral of (4x+e^x)sin(6x^2+3e^x)

\int (4 x + e^{x}) sin (6 x^{2} + 3 e^{x}) d x

- \frac{1}{3} cos (6 x^{2} + 3 e^{x}) + C

Solution steps

\int (4 x + e^{x}) sin (6 x^{2} + 3 e^{x}) d x

Apply u-substitution

: \int \frac{sin ( u )}{3} d u

= \int \frac{sin ( u )}{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 sin (u) d u

Use the common integral:

\int sin (u) d u = - cos (u)

= \frac{1}{3} (- cos (u))

Substitute back

u = 6 x^{2} + 3 e^{x}

= \frac{1}{3} (- cos (6 x^{2} + 3 e^{x}))

Simplify

= - \frac{1}{3} cos (6 x^{2} + 3 e^{x})

Add a constant to the solution

= - \frac{1}{3} cos (6 x^{2} + 3 e^{x}) + C

\int \frac{1 - x}{1 + x ^{2}} d x

\frac{d}{d x} ((x^{2} + 1)^{x + 1})

\frac{d}{d x} (c x + c - c^{2})

\int_{0}^{\infty} x \cdot e^{- x} d x

\frac{1}{2 x + 1}, at a = 3

What is the integral of (4x+e^x)sin(6x^2+3e^x) ?
The integral of (4x+e^x)sin(6x^2+3e^x) is -1/3 cos(6x^2+3e^x)+C