Integrate − 3 x 2 + 1 2 x + 3 2 cos ⁡ ( 2 x ) + 3 with respect to x.

Question

Integrate   −      3          x              2              +      1          2      x        +      3    2    cos  ⁡  (  2  x  )  +  3 with respect to x.

Kaylin Barry · Accepted Answer

Split the single integral into multiple integrals.

$\int - \frac{3}{x^{2}} d x + \int \frac{1}{2 x} d x + \int \frac{3}{2} \cdot \cos (2 x) d x + \int 3 d x$

Since $- 1$ is constant with respect to $x$ , move $- 1$ out of the integral.

$- \int \frac{3}{x^{2}} d x + \int \frac{1}{2 x} d x + \int \frac{3}{2} \cdot \cos (2 x) d x + \int 3 d x$

Since $3$ is constant with respect to $x$ , move $3$ out of the integral.

$- (3 \int \frac{1}{x^{2}} d x) + \int \frac{1}{2 x} d x + \int \frac{3}{2} \cdot \cos (2 x) d x + \int 3 d x$

Simplify the expression.

$- 3 \int x^{- 2} d x + \int \frac{1}{2 x} d x + \int \frac{3}{2} \cdot \cos (2 x) d x + \int 3 d x$

By the Power Rule, the integral of $x^{- 2}$ with respect to $x$ is $- x^{- 1}$ .

$- 3 (- x^{- 1} + C) + \int \frac{1}{2 x} d x + \int \frac{3}{2} \cdot \cos (2 x) d x + \int 3 d x$

Since $\frac{1}{2}$ is constant with respect to $x$ , move $\frac{1}{2}$ out of the integral.

$- 3 (- x^{- 1} + C) + \frac{1}{2} \int \frac{1}{x} d x + \int \frac{3}{2} \cdot \cos (2 x) d x + \int 3 d x$

The integral of $\frac{1}{x}$ with respect to $x$ is $\ln (| x |)$ .

$- 3 (- x^{- 1} + C) + \frac{1}{2} (\ln (| x |) + C) + \int \frac{3}{2} \cdot \cos (2 x) d x + \int 3 d x$

Since $\frac{3}{2}$ is constant with respect to $x$ , move $\frac{3}{2}$ out of the integral.

$- 3 (- x^{- 1} + C) + \frac{1}{2} (\ln (| x |) + C) + \frac{3}{2} \int \cos (2 x) d x + \int 3 d x$

Let $u = 2 x$ . Then $d u = 2 d x$ , so $\frac{1}{2} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

$- 3 (- x^{- 1} + C) + \frac{1}{2} (\ln (| x |) + C) + \frac{3}{2} \int \cos (u) \frac{1}{2} d u + \int 3 d x$

Combine $\cos (u)$ and $\frac{1}{2}$ .

$- 3 (- x^{- 1} + C) + \frac{1}{2} (\ln (| x |) + C) + \frac{3}{2} \int \frac{\cos (u)}{2} d u + \int 3 d x$

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

$- 3 (- x^{- 1} + C) + \frac{1}{2} (\ln (| x |) + C) + \frac{3}{2} (\frac{1}{2} \int \cos (u) d u) + \int 3 d x$

Simplify.

$- 3 (- x^{- 1} + C) + \frac{1}{2} (\ln (| x |) + C) + \frac{3}{4} \int \cos (u) d u + \int 3 d x$

The integral of $\cos (u)$ with respect to $u$ is $\sin (u)$ .

$- 3 (- x^{- 1} + C) + \frac{1}{2} (\ln (| x |) + C) + \frac{3}{4} (\sin (u) + C) + \int 3 d x$

Apply the constant rule.

$- 3 (- x^{- 1} + C) + \frac{1}{2} (\ln (| x |) + C) + \frac{3}{4} (\sin (u) + C) + 3 x + C$

Simplify.

$\frac{3}{x} + \frac{\ln (| x |)}{2} + \frac{3 \sin (u)}{4} + 3 x + C$

Replace all occurrences of $u$ with $2 x$ .

$\frac{3}{x} + \frac{\ln (| x |)}{2} + \frac{3 \sin (2 x)}{4} + 3 x + C$

Reorder terms.

$\frac{3}{x} + \frac{1}{2} \ln (| x |) + \frac{3}{4} \sin (2 x) + 3 x + C$

Integrate − 3 x 2 </mrow> + 1 <mrow

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