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

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

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integral of (2x+3)sin(x^2+3x+1)

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

- cos (x^{2} + 3 x + 1) + C

Solution steps

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

Apply u-substitution

: \int sin (u) d u

= \int sin (u) d u

Use the common integral:

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

= - cos (u)

Substitute back

u = x^{2} + 3 x + 1

= - cos (x^{2} + 3 x + 1)

Add a constant to the solution

= - cos (x^{2} + 3 x + 1) + C

\frac{\partial}{\partial z} (x e^{6 y} sin (5 z))

\int e^{x^{\frac{1}{2}}} d x

y = (4 x + 3)^{5}

x^{- \frac{1}{3}}

f (x) = x - (x + 6)^{6}

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