Integration Techniques
Example

Integrate

$$\int x^3 \ln(x) dx$$

A solution

Let $u = x^4$ so that $du = 4 x^3 dx$. Note that $4 \ln(x) = \ln(x^4)$. So,

$$\begin{eqnarray*} \int x^3 \ln(x) dx & = & \frac{1}{16} \int \ln(x^4) (4x^3) dx ... ... u) + C \\ & = & \frac{1}{4} x^4 \ln(x) - \frac{1}{16} x^4 + C \end{eqnarray*}$$

Another solution

Set $u = \ln(x)$ and $dv = x^3 dx$. So that $du = \frac{dx}{x}$ and $v = \frac{1}{4} x^4$.

Then,

$$\begin{eqnarray*} \int x^3 \ln(x) dx & = & \int u dv \\ & = & uv - \int v du \... ...x^3 dx \\ & = & \frac{1}{4} x^4 \ln(x) - \frac{1}{16} x^4 + C \end{eqnarray*}$$

Another example

Compute

$$\int \frac{4x}{\sqrt{x - 1}} dx$$

Solution

Set $u = 4x$ and $dv = \frac{dx}{\sqrt{x - 1}} = (x-1)^{-\frac{1}{2}} dx$ so that $du = 4dx$ and $v = 2 \sqrt{x-1}$. Then

$$\begin{eqnarray*} \int \frac{4x}{\sqrt{x - 1}} dx & = & \int u dv \\ & = & 8x ... ...\ & = & 8x \sqrt{x - 1} - \frac{16}{3} (x - 1)^\frac{3}{2} + C \end{eqnarray*}$$

Another example

Compute

$$\int \frac{\ln(\ln(x))}{x \ln(x)} dx$$

Solution

Set $u = \ln(\ln(x))$ so that $du = \frac{dx}{x \ln(x)}$.

Then

$$\begin{eqnarray*} \int \frac{\ln(\ln(x))}{x \ln(x)} dx & = & \int du \\ & = & u + C \\ & = & \ln(\ln(x)) + C \end{eqnarray*}$$

Another example

Compute

$$\int_{-\pi}^\pi x \sin(x^4) dx$$

Solution

If $f(x) = x \sin(x^4)$, then $f(-x) = -f(x)$. Thus, $\int_{-\pi}^0 x \sin(x^4) dx = - \int_0^\pi x \sin(x^4) dx$. So, $\int_{-\pi}^\pi x \sin(x^4) dx = 0$.