\begin{align*} \frac{\dif }{\dif x}\left(f(x)g(x)\right) &= f’(x)g(x) + f(x)g’(x)\\ \int\frac{\dif }{\dif x}\left(f(x)g(x)\right)\dif x &= \int \left(f’(x)g(x) + f(x)g’(x)\right)\dif x\\ f(x)g(x) + c &= \int f’(x)g(x) \dif x + \int f(x)g’(x)\dif x\\ \int f’(x)g(x) \dif x &= f(x)g(x) + c - \int f(x)g’(x)\dif x \end{align*} When we integrate $f(x)g’(x)$ we’ll also get a constant, so the $+c$ after $f(x)g(x)$ is unnecessary, and we can drop it.

A slightly different way to derive this formula would be to re-arrange the product rule first to get $f’(x)g(x) = \frac{\dif }{\dif x}\left(f(x)g(x)\right) - f(x)g’(x)$ and then integrate.

If we choose $\ln x$ as the part to integrate we quickly run into trouble, because at this stage we don’t know how to integrate $\ln x$ (see Question 15.4.23 for that). So we must integrate the $x$ part, and differentiate the $\ln x$ part: \begin{align*} \int x\ln x\dif x &= \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \frac{1}{x} \dif x\\ &= \frac{x^2}{2} \ln x - \int \frac{x}{2} \dif x\\ &= \frac{x^2}{2} \ln x - \frac{x^2}{4} + c \end{align*}

Choosing to integrate the $e^x$ part and differentiate the $x$ part gives $1e^x$ which we can integrate easily, whereas the other choice gives $f(x)g’(x) = \frac{1}{2}x^2 e^x$ which is even worse than the original function. \begin{align*} \int_0^2 x e^x \dif x &= \left.x e^x\right|_0^2 - \int_0^2 1 e^x \dif x\\ &= 2e^2 - 0e^0 - \left[ e^x \right]_0^2\\ &= 2e^2 - \left[e^2 - e^0\right]\\ &= e^2 + 1. \end{align*}

\begin{align*} \int\ln x\dif x &= \int1\times\ln x\dif x\\ &= x\ln x - \int x\times \frac{1}{x}\dif x\\ &= x\ln x - \int 1\dif x\\ &= x\ln x - x + c \end{align*} To check that this is correct, you can differentiate it: \begin{align*} \frac{\dif }{\dif x} (x\ln x - x + c) &= \left(1\ln x + x\frac{1}{x}\right) - 1 + 0\\ &=\ln x \end{align*}

We’ll have to use integration by parts twice: \begin{align*} \int x^2\cos x\dif x &= x^2\sin x - \int 2x\sin x\dif x\\ &= x^2\sin x - \left(2x(-\cos x) - \int 2(-\cos x)\dif x \right)\\ &= x^2\sin x + 2x\cos x - 2\sin x + c \end{align*} You can check your answer by differentiating it: \begin{align*} \frac{\dif }{\dif x} \left(x^2\sin x + 2x\cos x - 2\sin x + c\right)&= \left(2x\sin x + x^2\cos x\right) + \left(2\cos x - 2x\sin x\right) - 2\cos x + 0\\ &=x^2\cos x \end{align*}

There are a few ways to do this one. We can choose $e^x$ as the function to integrate: \begin{align*} \int\e^{x}\cos x\dif x &= e^x\cos x - \int e^x(-\sin x)\dif x\\ &= e^x\cos x + \int e^x\sin x\dif x\\ &= e^x\cos x + e^x\sin x - \int e^x \cos x \dif x \end{align*} We end up with $\int e^x \cos x \dif x$ again, but that’s ok because we can just solve for it: \begin{align*} 2\int e^x \cos x \dif x &= e^x\cos x + e^x\sin x\\ \int e^x \cos x \dif x &= \frac{e^x\cos x + e^x\sin x}{2} + c \end{align*} (We need to add in the $c$ at the end because of the $c$ we got rid of when we were simplifying the formula for integration by parts in Question 15.4.20.) If it makes things clearer for you, you can give the integral a variable, e.g. let $I = \int e^x \cos x \dif x$, then \begin{align*} I &= e^x\cos x + e^x\sin x - I\\ 2I &= e^x\cos x + e^x\sin x\\ I &= \frac{e^x\cos x + e^x\sin x}{2} \end{align*}

We could instead choose $\cos x$ as the function to integrate: \begin{align*} \int\e^{x}\cos x\dif x &= e^x \sin x - \int e^x \sin x\dif x\\ &= e^x\sin x - \left(e^x(-\cos x) - \int e^x(-\cos x)\dif x\right)\\ &= e^x\sin x + e^x\cos x - \int\e^{x}\cos x\dif x \end{align*} and proceed as before.

A third way to solve this would be to do the integration once each way: \begin{align*} \int\e^{x}\cos x\dif x &= e^x\cos x - \int e^x(-\sin x)\dif x \\ &= e^x\cos x + \int e^x\sin x\dif x \tag{$1$}\\ \int\e^{x}\cos x\dif x &= e^x \sin x - \int e^x \sin x\dif x \tag{$2$} \end{align*} and then solve these two equations simultaneously: \begin{align*} (1) + (2):&&2\int\e^{x}\cos x\dif x &= e^x\cos x + e^x \sin x \end{align*} and so on as before.

When we have a complicated function like $\e^{\sqrt{x}}$ then it’s usually best to do a substitution, so let $u = \sqrt{x}$, then \begin{align*} \frac{\dif u}{\dif x} &= \frac{1}{2\sqrt{x}}\\ \dif x &= 2\sqrt{x}\dif u\\ &= 2u \dif u \end{align*} \begin{align*} \int \e^{\sqrt{x}}\dif x &= \int \e^u 2u\dif u \end{align*} Now we can apply integration by parts, integrating $\e^u$ and differentiating $2u$. (Or you might decide to bring the $2$ out the front of the integral first.) \begin{align*} \int \e^{\sqrt{x}}\dif x &= \e^u 2u - \int 2\e^u \dif u\\ &= 2u\e^u - 2\e^u + c\\ &= 2\e^u(u - 1) + c\\ &= 2\e^{\sqrt{x}}(\sqrt{x} - 1) + c \end{align*}