evaluate the following indefinite integral. ∫(π/x⁴ + e√x) dx ∫(π/x⁴ + e√x) dx = (type an exact answer.)

Explanation:

Step1: Split the integral

By the sum - rule of integration $\int(f(x)+g(x))dx=\int f(x)dx+\int g(x)dx$, we have $\int(\frac{\pi}{x^{4}}+e\sqrt{x})dx=\int\frac{\pi}{x^{4}}dx+\int e\sqrt{x}dx$.

Step2: Rewrite the functions

Rewrite $\frac{\pi}{x^{4}}$ as $\pi x^{- 4}$ and $\sqrt{x}$ as $x^{\frac{1}{2}}$. So the integral becomes $\pi\int x^{-4}dx+e\int x^{\frac{1}{2}}dx$.

Step3: Apply the power - rule of integration

The power - rule is $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$). For $\int x^{-4}dx$, we have $\frac{x^{-4 + 1}}{-4+1}=-\frac{1}{3}x^{-3}$. For $\int x^{\frac{1}{2}}dx$, we have $\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}=\frac{2}{3}x^{\frac{3}{2}}$.

Step4: Combine the results

$\pi\int x^{-4}dx+e\int x^{\frac{1}{2}}dx=\pi(-\frac{1}{3}x^{-3})+e(\frac{2}{3}x^{\frac{3}{2}})+C=-\frac{\pi}{3x^{3}}+\frac{2e}{3}x^{\frac{3}{2}}+C$.

Answer:

$-\frac{\pi}{3x^{3}}+\frac{2e}{3}x^{\frac{3}{2}}+C$