Questions: ∫(3x^4 + 2/x - 3/x^5 - 6 √x) dx

Solution Steps

To solve the integral \(\int\left(3 x^{4}+\frac{2}{x}-\frac{3}{x^{5}}-6 \sqrt{x}\right) d x\), we will integrate each term separately. The integral of a sum is the sum of the integrals. We will use the power rule for integration, which states that \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\) for any real number \(n \neq -1\).

1. Integrate \(3x^4\) using the power rule.
2. Integrate \(\frac{2}{x}\) which is equivalent to \(2x^{-1}\).
3. Integrate \(-\frac{3}{x^5}\) which is equivalent to \(-3x^{-5}\).
4. Integrate \(-6\sqrt{x}\) which is equivalent to \(-6x^{1/2}\).
5. Sum all the integrals and add the constant of integration \(C\).

We start with the integral

\[ \int\left(3 x^{4}+\frac{2}{x}-\frac{3}{x^{5}}-6 \sqrt{x}\right) d x. \]

We will integrate each term separately.

Using the power rule, we have:

\[ \int 3x^4 \, dx = \frac{3}{5}x^5. \]

The integral of \(\frac{2}{x}\) is:

\[ \int \frac{2}{x} \, dx = 2 \ln |x|. \]

For \(-\frac{3}{x^5}\), we apply the power rule:

\[ \int -\frac{3}{x^5} \, dx = \frac{3}{4x^4}. \]

The integral of \(-6\sqrt{x}\) is:

\[ \int -6\sqrt{x} \, dx = -4x^{3/2}. \]

Combining all the results, we have:

\[ \int\left(3 x^{4}+\frac{2}{x}-\frac{3}{x^{5}}-6 \sqrt{x}\right) d x = -4x^{3/2} + \frac{3}{5}x^5 + 2 \ln |x| + \frac{3}{4x^4} + C, \]

where \(C\) is the constant of integration.

Final Answer