The problem is to evaluate the following four integrals, showing two different methods for solving each one. Integral 1: $\int x^2(x^3+1)^2 dx$ Integral 2: $\int e^{\sqrt{5}+e^x} dx$ Integral 3: $\int \sin(\frac{x}{4}) \cos(\frac{x}{4}) dx$ Integral 4: $\int \frac{dx}{x-\sqrt{x}}$

1. Problem Description

The problem is to evaluate the following four integrals, showing two different methods for solving each one.

Integral 1: $\int x^2(x^3+1)^2 dx$

Integral 2: $\int e^{\sqrt{5}+e^x} dx$

Integral 3: $\int \sin(\frac{x}{4}) \cos(\frac{x}{4}) dx$

Integral 4: $\int \frac{dx}{x-\sqrt{x}}$

2. Solution Steps

Integral 1: $\int x^2(x^3+1)^2 dx$

Method 1: Expand and integrate term by term.

$\int x^2(x^3+1)^2 dx = \int x^2(x^6+2x^3+1) dx = \int (x^8+2x^5+x^2) dx = \frac{x^9}{9} + \frac{2x^6}{6} + \frac{x^3}{3} + C = \frac{x^9}{9} + \frac{x^6}{3} + \frac{x^3}{3} + C$

Method 2: u-substitution. Let $u = x^3+1$, then $du = 3x^2 dx$, so $x^2 dx = \frac{1}{3} du$.

$\int x^2(x^3+1)^2 dx = \int u^2 \frac{1}{3} du = \frac{1}{3} \int u^2 du = \frac{1}{3} \frac{u^3}{3} + C = \frac{1}{9} (x^3+1)^3 + C = \frac{1}{9} (x^9 + 3x^6 + 3x^3 + 1) + C = \frac{x^9}{9} + \frac{x^6}{3} + \frac{x^3}{3} + \frac{1}{9} + C = \frac{x^9}{9} + \frac{x^6}{3} + \frac{x^3}{3} + C'$

(where $C' = \frac{1}{9} + C$)

Integral 2: $\int e^{\sqrt{5}+e^x} dx$

Method 1: Rewrite the integral.

$\int e^{\sqrt{5}+e^x} dx = \int e^{\sqrt{5}} e^{e^x} dx = e^{\sqrt{5}} \int e^{e^x} dx$.

Let $u=e^x$, then $du=e^x dx$, so $dx=\frac{du}{u}$.

$e^{\sqrt{5}} \int \frac{e^u}{u} du$.

The integral $\int \frac{e^u}{u} du$ cannot be expressed in terms of elementary functions. It is an exponential integral denoted as $Ei(u)$.

So, the solution is $e^{\sqrt{5}}Ei(e^x)+C$.

Method 2: Rewriting $e^{\sqrt{5}+e^x}$ as $e^{\sqrt{5}}e^{e^x}$ and then attempting integration by parts will not work.

Use the exponential integral directly. Since $\int \frac{e^u}{u}du = Ei(u)+C$, then $\int e^{e^x} dx = Ei(e^x)+C$.

Therefore, $\int e^{\sqrt{5}+e^x} dx = e^{\sqrt{5}}\int e^{e^x} dx = e^{\sqrt{5}}Ei(e^x) + C$.

Integral 3: $\int \sin(\frac{x}{4}) \cos(\frac{x}{4}) dx$

Method 1: Use the trigonometric identity $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$, so $\sin(\theta)\cos(\theta) = \frac{1}{2}\sin(2\theta)$.

$\int \sin(\frac{x}{4}) \cos(\frac{x}{4}) dx = \int \frac{1}{2} \sin(\frac{2x}{4}) dx = \frac{1}{2} \int \sin(\frac{x}{2}) dx = \frac{1}{2} (-2\cos(\frac{x}{2})) + C = -\cos(\frac{x}{2}) + C$

Method 2: u-substitution. Let $u = \sin(\frac{x}{4})$. Then $du = \frac{1}{4} \cos(\frac{x}{4}) dx$.

$\int \sin(\frac{x}{4}) \cos(\frac{x}{4}) dx = \int u \cdot 4 du = 4 \int u du = 4 \frac{u^2}{2} + C = 2u^2 + C = 2\sin^2(\frac{x}{4}) + C$

We need to verify if $-\cos(\frac{x}{2}) + C = 2\sin^2(\frac{x}{4}) + C'$.

Using the half-angle identity $\cos(x) = 1 - 2\sin^2(\frac{x}{2})$, we have $\cos(\frac{x}{2}) = 1 - 2\sin^2(\frac{x}{4})$.

So, $-\cos(\frac{x}{2}) + C = -(1 - 2\sin^2(\frac{x}{4})) + C = -1 + 2\sin^2(\frac{x}{4}) + C = 2\sin^2(\frac{x}{4}) + (C-1)$. Therefore, $C' = C-1$.

Integral 4: $\int \frac{dx}{x-\sqrt{x}}$

Method 1: u-substitution. Let $u = \sqrt{x}$. Then $u^2 = x$, so $2u du = dx$.

$\int \frac{dx}{x-\sqrt{x}} = \int \frac{2u du}{u^2-u} = \int \frac{2u du}{u(u-1)} = \int \frac{2}{u-1} du = 2 \ln|u-1| + C = 2 \ln|\sqrt{x}-1| + C$

Method 2: Rewriting the integral

$\int \frac{dx}{x - \sqrt{x}} = \int \frac{dx}{\sqrt{x}(\sqrt{x}-1)}$.

Let $u=\sqrt{x}-1$, then $du = \frac{1}{2\sqrt{x}}dx$. This can be rewritten as $dx = 2\sqrt{x}du$.

Therefore,

$\int \frac{dx}{\sqrt{x}(\sqrt{x}-1)} = \int \frac{2\sqrt{x}du}{\sqrt{x}u} = \int \frac{2}{u}du = 2\ln|u| + C = 2\ln|\sqrt{x}-1| + C$.

3. Final Answer

Integral 1: $\int x^2(x^3+1)^2 dx = \frac{x^9}{9} + \frac{x^6}{3} + \frac{x^3}{3} + C$ (Method 1) or $\frac{1}{9} (x^3+1)^3 + C$ (Method 2)

Integral 2: $\int e^{\sqrt{5}+e^x} dx = e^{\sqrt{5}}Ei(e^x) + C$ (Method 1 and 2). $Ei(x)$ is the exponential integral function.

Integral 3: $\int \sin(\frac{x}{4}) \cos(\frac{x}{4}) dx = -\cos(\frac{x}{2}) + C$ (Method 1) or $2\sin^2(\frac{x}{4}) + C$ (Method 2)

Integral 4: $\int \frac{dx}{x-\sqrt{x}} = 2 \ln|\sqrt{x}-1| + C$ (Method 1 and 2)