What is the integral of (6x+3)cos(x^2+x) ?

The integral of (6x+3)cos(x^2+x) is 3sin(x^2+x)+C

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integral of (6x+3)cos(x^2+x)

\int (6 x + 3) cos (x^{2} + x) d x

3 sin (x^{2} + x) + C

Solution steps

\int (6 x + 3) cos (x^{2} + x) d x

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

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

Take the constant out:

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

= 3 \cdot \int (2 x + 1) cos (x^{2} + x) d x

Apply u-substitution

: \int cos (u) d u

= 3 \cdot \int cos (u) d u

Use the common integral:

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

= 3 sin (u)

Substitute back

u = x^{2} + x

= 3 sin (x^{2} + x)

Add a constant to the solution

= 3 sin (x^{2} + x) + C

\frac{\partial}{\partial y} (\frac{1}{e ^{x^{2} + y^{2}}})

\int \frac{x ^{3}}{( 2 + 3 x ^{4} ) ^{3}} d x

n = 1 \sum \infty \frac{7}{n}

\frac{\partial}{\partial x} (3 - 3 x^{2} + y^{2})

(8 + x) y^{^{'}} = 5 y

What is the integral of (6x+3)cos(x^2+x) ?
The integral of (6x+3)cos(x^2+x) is 3sin(x^2+x)+C