Integration by Parts

Integration by parts: $\displaystyle\int u\,dv = uv - \int v\,du$. Choose $u$ as the function that appears EARLIEST in LIATE (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential) — it differentiates toward something simpler, while the rest is $dv$. A lone $\log x$ or $\tan^{-1}x$ is integrated by taking $dv = dx$.

Apply integration by parts twice. The same integral $I$ reappears on the right with a coefficient; collect it: $I = (\text{boundary terms}) + kI\Rightarrow I(1-k) = \text{terms}$. For $\int \sin(\log x)\,dx$, the substitution $x = e^t$ turns it into $\int e^t \sin t\,dt$, the classic cyclic form.

$\displaystyle\int e^x\big[f(x) + f'(x)\big]\,dx = e^x f(x) + C$. The work is REWRITING the integrand into this shape — using identities so that one part is a function $f$ and the rest is exactly its derivative $f'$. A close cousin: $\displaystyle\int e^x\dfrac{1 + x\log x}{x}$-style problems all reduce to spotting $f + f'$.

The three standard results (each derived by taking $\sqrt{\,\cdot\,}$ as the first function and $1$ as the second):

• $\displaystyle\int \sqrt{a^2 - x^2}\,dx = \dfrac{x}{2}\sqrt{a^2 - x^2} + \dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a} + C$
• $\displaystyle\int \sqrt{a^2 + x^2}\,dx = \dfrac{x}{2}\sqrt{a^2 + x^2} + \dfrac{a^2}{2}\log\left|x + \sqrt{a^2 + x^2}\right| + C$
• $\displaystyle\int \sqrt{x^2 - a^2}\,dx = \dfrac{x}{2}\sqrt{x^2 - a^2} - \dfrac{a^2}{2}\log\left|x + \sqrt{x^2 - a^2}\right| + C$

For a general $\sqrt{ax^2+bx+c}$, complete the square first, then match one of these three.

If $u$ is a polynomial (so some derivative $u^{(n)} = 0$) and $v$ the second function, then $\displaystyle\int u\,v\,dx = u\,v_1 - u'\,v_2 + u''\,v_3 - u'''\,v_4 + \cdots$, where a prime is a derivative of $u$ and a subscript $k$ means integrate $v$ $k$ times. The series terminates because the polynomial's derivatives reach zero.