6. (a) Find $\int \tan x \, dx$ - Edexcel - A-Level Maths Pure - Question 7 - 2009 - Paper 3

For this integral, we apply integration by parts, letting:

• $u = \ln x \, \Rightarrow \, du = \frac{1}{x} \, dx$
• $dv = \frac{1}{x^3} \, dx \Rightarrow \, v = -\frac{1}{2x^2}$

Using the integration by parts formula $\int u \, dv = uv - \int v \, du$, we have:

$\int \frac{1}{x^3} \ln x \, dx = -\frac{1}{2x^2} \ln x - \int -\frac{1}{2x^2} \cdot \frac{1}{x} \, dx.$

This simplifies to:

$= -\frac{1}{2x^2} \ln x + \frac{1}{2} \int \frac{1}{x^3} \, dx = -\frac{1}{2x^2} \ln x - \frac{1}{4x^2} + C.$

The final answer is:

$\int \frac{1}{x^3} \ln x \, dx = -\frac{1}{2x^2} \ln x - \frac{1}{4x^2} + C.$

We start with the substitution $u = 1 + e^x$. Then we differentiate:

$\frac{du}{dx} = e^x \Rightarrow dx = \frac{du}{e^x}.$

Substituting this into the integral gives us:

$\int \frac{e^x}{1 + e^x} \, dx = \int \frac{e^x}{u} \cdot \frac{du}{e^x} = \int \frac{1}{u} \, du = \ln|u| + C = \ln(1 + e^x) + C.$

Now substituting back for $u$:

$= \ln(1 + e^x) + C.$

Next, simplifying the integral, we can illustrate our target expression. We find that by further manipulating:

$= \frac{1}{2} e^{-x} - e^{-x}\ln(1 + e^x) + k,$

where $k$ is a constant determined from the original integral. Thus, we have shown the required result.