# Find the integral. \int (2x^5 - 7x^3 + 5)\ dx

## Question:

Find the integral.

{eq}\int (2x^5 - 7x^3 + 5)\ dx {/eq}

## Rules for Evaluating Indefinite Integrals:

The indefinite integral operation obeys the following rules:

1) The linearity rule:

{eq}\displaystyle \int \left[af(x)+bg(x)\right] \, dx = a\int f(x) \, dx + b\int g(x) \, dx \, . {/eq}

2) The power rule: if {eq}a \ne -1 {/eq}, then

{eq}\displaystyle \int x^a \, dx = \frac{1}{a+1}x^{a+1} + C \, . {/eq}

3) Integration by substitution:

{eq}\displaystyle \int f(g(x)) g'(x) \, dx = \int f(u) \, du \, . {/eq}

4) Integration by parts:

{eq}\displaystyle \int u(x) v'(x) \, dx = u(x)v(x) - \int v(x) u'(x) \, dx \, . {/eq}

We can use the linearity and power rules for indefinite integrals to evaluate the given integral. We have:

{eq}\begin{align*} \int (2x^5-7x^3+5) \, dx &= 2\int x^5 \, dx-7\int x^3 \, dx+5 \int dx&&\text{(by the linearity rule)}\\ &=2\left(\frac{1}{6}x^6+C_1\right)-7\left(\frac{1}{4}x^4+C_2\right)+5(x+C_3)&&\text{(evaluating each individual integral by the power rule; }C_1,C_2,C_3\text{ are constants of integration)}\\ &=\frac{1}{3}x^6-\frac{7}{4}x^4+5x+2C_1-7C_2+5C_3&&\text{(expanding, rearranging)}\\ &=\frac{1}{3}x^6-\frac{7}{4}x^4+5x+C&&\text{(combining all constants of integration into a single arbitrary constant }C\text{).} \end{align*} {/eq}

In summary, {eq}\boxed{\int (2x^5-7x^3+5) \, dx = \frac{1}{3}x^6-\frac{7}{4}x^4+5x+C \, .} {/eq}