Question

calculate the integral and write the answer in simplest form.

$$ int left( 3 x ^ { - 3 } + 3 x ^ { 5 } + 4 ight) d x $$

Explanation:

Step1: Apply the sum rule of integration

$$\int(3x^{-3}+3x^{5}+4)dx=\int3x^{-3}dx+\int3x^{5}dx+\int4dx$$

Step2: Use the constant multiple rule

$$=3\int x^{-3}dx + 3\int x^{5}dx+4\int dx$$

Step3: Apply the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n

eq - 1)$ and $\int dx=x + C$
For $\int x^{-3}dx$, $n=-3$, so $\int x^{-3}dx=\frac{x^{-3 + 1}}{-3+1}=\frac{x^{-2}}{-2}=-\frac{1}{2x^{2}}$
For $\int x^{5}dx$, $n = 5$, so $\int x^{5}dx=\frac{x^{5+1}}{5+1}=\frac{x^{6}}{6}$
For $\int dx$, we have $\int dx=x$

Then $3\int x^{-3}dx+3\int x^{5}dx + 4\int dx=3\times(-\frac{1}{2x^{2}})+3\times\frac{x^{6}}{6}+4x+C$

Step4: Simplify the expression

$$=-\frac{3}{2x^{2}}+\frac{x^{6}}{2}+4x+C$$

Answer:

$-\frac{3}{2x^{2}}+\frac{x^{6}}{2}+4x+C$