What is the integral of (15x^{-2}-5x)cos(6x^{-1}+x^2) ?

The integral of (15x^{-2}-5x)cos(6x^{-1}+x^2) is -5/2 sin(6/x+x^2)+C

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integral of (15x^{-2}-5x)cos(6x^{-1}+x^2)

\int (15 x^{- 2} - 5 x) cos (6 x^{- 1} + x^{2}) d x

- \frac{5}{2} sin (\frac{6}{x} + x^{2}) + C

Solution steps

\int (15 x^{- 2} - 5 x) cos (6 x^{- 1} + x^{2}) d x

Apply u-substitution

: \int - \frac{5 cos ( u )}{2} d u

= \int - \frac{5 cos ( u )}{2} d u

Take the constant out:

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

= - \frac{5}{2} \cdot \int cos (u) d u

Use the common integral:

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

= - \frac{5}{2} sin (u)

Substitute back

u = 6 x^{- 1} + x^{2}

= - \frac{5}{2} sin (6 x^{- 1} + x^{2})

6 x^{- 1} = \frac{6}{x}

= - \frac{5}{2} sin (\frac{6}{x} + x^{2})

Add a constant to the solution

= - \frac{5}{2} sin (\frac{6}{x} + x^{2}) + C

\int 2^{s} d s

\frac{d y}{d x} = y - 3

y^{^{''}} - 10 y^{^{'}} + y = 0

\int \frac{24 y - 14}{6 y ^{2} - 7 y} d y

\frac{1}{s ( s + 1 )} - \frac{1}{s + 1}

What is the integral of (15x^{-2}-5x)cos(6x^{-1}+x^2) ?
The integral of (15x^{-2}-5x)cos(6x^{-1}+x^2) is -5/2 sin(6/x+x^2)+C