Prelude

The easy integral was really hard for me since I didn’t know the standard form (not taught in school), and couldn’t integrate $sec\theta$ either

Beginner Integral

\begin{gathered} I = \int_2^3 x(x-2)^4 \ dx \\ \text{Let $u = x-2$, $dx = du$} \\ I = \int_0^1 u^4(u+2)\ du \\ = \int_0^1 u^5 + 2u^4\ du \\ = \left[\frac{1}{6}u^6 + \frac{2}{5}u^5\right]^1_0 \\ = \frac{1}{6} + \frac{2}{5} \\ = \frac{17}{30} \end{gathered}

\begin{gathered} \int_1^2 \frac{\sqrt{x^2+1}-\sqrt{x^2-1}}{\sqrt{x^4-1}} \ dx \\ = \int_1^2 \frac{\sqrt{x^2+1}}{\sqrt{x^4-1}} - \frac{\sqrt{x^2-1}}{\sqrt{x^4-1}} \ dx \\ = \int_1^2 \frac{1}{\sqrt{x^2-1}} - \frac{1}{\sqrt{x^2+1}}\ dx\\ \text{Let $I_1 = \int_1^2 \frac{1}{\sqrt{x^2-1}}\ dx$} \\ \text{Let $x = \sec \theta$, $dx = \sec \theta\ \tan \theta\ d\theta$} \\ I_1 = \int_0^\frac{\pi}{3} \frac{\sec \theta\ \tan \theta}{\sqrt{\sec^2 \theta - 1}}\ d\theta \\ = \int_0^\frac{\pi}{3} \frac{\sec \theta\ \tan \theta}{\tan \theta}\ d\theta \\ = \int_0^\frac{\pi}{3} \sec \theta\ d\theta = \left[\ln\left|\sec \theta+\tan \theta\right|\right]_0^\frac{\pi}{3} \\ =\ln(2+\sqrt 3) \\ \ \\ \text{Let $I_2 = \int_1^2 \frac{1}{\sqrt{x^2+1}}\ dx$} \\ \text{Let $x = \tan\theta$, $dx = \sec^2\theta\ d\theta$} \\ I_2 = \int_\frac{\pi}{4}^{\arctan{2}} \frac{\sec^2\theta}{\sqrt{\tan^2\theta + 1}} \ d\theta \\ = \int_\frac{\pi}{4}^{\arctan{2}} \frac{\sec^2\theta}{\sec\theta}\ d\theta \\ = \int_\frac{\pi}{4}^{\arctan{2}} \sec\theta\ d\theta \\ = \left[\ln\left|\sec\theta + \tan\theta\right|\right]_\frac{\pi}{4}^{\arctan{2}} \\ = \ln(\sqrt 5 + 2) - \ln(\sqrt 2 + 1) \\ \ \\ I_1 - I_2 = \boxed{\ln(2 + \sqrt 3) - \ln(\sqrt 5 + 2) + \ln(\sqrt 2 + 1)} \end{gathered}

\begin{gathered} I = \int_0^1 xe^x\sin(x^2)\ dx \\ \text{Let $u = x^2$, $dx = 2x\ du$} \\ I = \frac{1}{2} \int_0^1e^u\sin u\ du \\ \text{Let $t = \sin u$, $dt = \cos u$, $dm = e^u = m$} \\ I = \frac{1}{2}\left(e^u\sin u - \int_0^1 e^u\cos u\ du\right) \\ \text{Let $t = \cos u$, $dt = -\sin u$, $dk = e^u = k$} \\ I = \frac{1}{2}\left(e^u\sin u - e^u\cos u - \int_0^1 e^u\sin u\ du\right) \\ 2I = e^u\sin u - e^u\cos u \\ I = \frac{1}{4}\left[e^u\sin u - e^u\cos u\right]_0^1 \\ =\boxed{\frac{1}{4}\left(e\sin 1 - e\cos 1 + 1\right)} \end{gathered}