Integration by Parts

$u = x$, $dv = e^{3x}\,dx$.

$du = dx$, $v = \tfrac13 e^{3x}$.

$\int xe^{3x}\,dx = \tfrac{x}{3}e^{3x} - \int\tfrac13 e^{3x}\,dx$.

No $x$ is left: $\int\tfrac13 e^{3x}\,dx = \tfrac19 e^{3x}$. One round is enough.

$\tfrac{x}{3}e^{3x} - \tfrac19 e^{3x} + C = \tfrac{e^{3x}}{9}(3x - 1) + C$.

$u = x$, $dv = \cos(2x)\,dx$.

$du = dx$, $v = \tfrac12\sin(2x)$.

$\Big[\tfrac{x}{2}\sin(2x)\Big]_0^{\pi/2} - \int_0^{\pi/2}\tfrac12\sin(2x)\,dx$.

$\int_0^{\pi/2}\tfrac12\sin(2x)\,dx = \Big[-\tfrac14\cos(2x)\Big]_0^{\pi/2} = \tfrac14 + \tfrac14 = \tfrac12$. No $x$ left.

$\Big[\tfrac{x}{2}\sin(2x)\Big]_0^{\pi/2} = \tfrac{\pi}{4}\sin\pi - 0 = 0$, so the total is $0 - \tfrac12 = -\tfrac12$.

$u = x^2$, $dv = e^{-x}\,dx$.

$du = 2x\,dx$, $v = -e^{-x}$.

$\int x^2e^{-x}\,dx = -x^2e^{-x} + \int 2xe^{-x}\,dx$.

Again with $u = 2x$, $dv = e^{-x}\,dx$: $\int 2xe^{-x}\,dx = -2xe^{-x} + \int 2e^{-x}\,dx = -2xe^{-x} - 2e^{-x}$.

$-x^2e^{-x} - 2xe^{-x} - 2e^{-x} + C = -e^{-x}(x^2 + 2x + 2) + C$.

First $w = \sqrt{x}$, so $x = w^2$, $dx = 2w\,dw$: $\int 2w\sin w\,dw$. Now $u = 2w$, $dv = \sin w\,dw$.

$du = 2\,dw$, $v = -\cos w$.

$\int 2w\sin w\,dw = -2w\cos w + \int 2\cos w\,dw$.

$\int 2\cos w\,dw = 2\sin w$. No $w$ left.

$-2w\cos w + 2\sin w + C = 2\sin\sqrt{x} - 2\sqrt{x}\cos\sqrt{x} + C$.

$u = \log x$, $dv = x^{1/2}\,dx$.

$du = \tfrac1x\,dx$, $v = \tfrac23 x^{3/2}$.

$\int\sqrt{x}\log x\,dx = \tfrac23 x^{3/2}\log x - \int\tfrac23 x^{3/2}\cdot\tfrac1x\,dx$.

$\int\tfrac23 x^{1/2}\,dx = \tfrac49 x^{3/2}$. Total: $\tfrac23 x^{3/2}\log x - \tfrac49 x^{3/2} + C$.

$u = \arctan(3x)$, $dv = dx$.

$du = \dfrac{3}{1 + 9x^2}\,dx$, $v = x$.

$x\arctan(3x) - \int\dfrac{3x}{1 + 9x^2}\,dx$.

$w = 1 + 9x^2$, $dw = 18x\,dx$: $\int\dfrac{3x}{1 + 9x^2}\,dx = \tfrac16\log(1 + 9x^2)$. Total: $x\arctan(3x) - \tfrac16\log(1 + 9x^2) + C$.

$u = \log x$, $dv = x^{-2}\,dx$.

$du = \tfrac1x\,dx$, $v = -\tfrac1x$.

$\Big[-\dfrac{\log x}{x}\Big]_1^e + \int_1^e\dfrac{1}{x^2}\,dx$.

$\Big[-\dfrac{\log x}{x}\Big]_1^e = -\dfrac1e$ and $\int_1^e x^{-2}\,dx = 1 - \dfrac1e$. Total: $1 - \dfrac2e$.

$I = \int e^{2x}\sin x\,dx$. $u = \sin x$, $dv = e^{2x}\,dx$, so $du = \cos x\,dx$, $v = \tfrac12 e^{2x}$.

$I = \tfrac12 e^{2x}\sin x - \tfrac12\int e^{2x}\cos x\,dx$.

$u = \cos x$, $dv = e^{2x}\,dx$: $\int e^{2x}\cos x\,dx = \tfrac12 e^{2x}\cos x + \tfrac12 I$.

$I = \tfrac12 e^{2x}\sin x - \tfrac14 e^{2x}\cos x - \tfrac14 I$, so $\tfrac54 I = \tfrac14 e^{2x}(2\sin x - \cos x)$. Divide by $\tfrac54$ and add $C$.

$I = \int_0^{\pi}e^{x}\cos x\,dx$. $u = \cos x$, $dv = e^{x}\,dx$, so $du = -\sin x\,dx$, $v = e^{x}$.

$I = \Big[e^{x}\cos x\Big]_0^{\pi} + \int_0^{\pi}e^{x}\sin x\,dx = -e^{\pi} - 1 + \int_0^{\pi}e^{x}\sin x\,dx$.

$u = \sin x$: $\int_0^{\pi}e^{x}\sin x\,dx = \Big[e^{x}\sin x\Big]_0^{\pi} - I = 0 - I$.

$I = -e^{\pi} - 1 - I$, so $2I = -(e^{\pi} + 1)$ and $I = -\dfrac{e^{\pi} + 1}{2}$.

$I = \int x\cos(\log x)\,dx$. $u = \cos(\log x)$, $dv = x\,dx$, so $du = -\dfrac{\sin(\log x)}{x}\,dx$, $v = \dfrac{x^2}{2}$.

$I = \dfrac{x^2}{2}\cos(\log x) + \dfrac12\int x\sin(\log x)\,dx$.

$u = \sin(\log x)$, $dv = x\,dx$: $\int x\sin(\log x)\,dx = \dfrac{x^2}{2}\sin(\log x) - \dfrac12 I$.

$I = \dfrac{x^2}{2}\cos(\log x) + \dfrac{x^2}{4}\sin(\log x) - \dfrac14 I$, so $\dfrac54 I = \dfrac{x^2}{4}\big(2\cos(\log x) + \sin(\log x)\big)$. Divide by $\dfrac54$ and add $C$.

$I_n = \int(\log x)^n\,dx$. $u = (\log x)^n$, $dv = dx$.

$du = \dfrac{n(\log x)^{n-1}}{x}\,dx$, $v = x$: $I_n = x(\log x)^n - \int x\cdot\dfrac{n(\log x)^{n-1}}{x}\,dx$.

The $x$ cancels: $I_n = x(\log x)^n - nI_{n-1}$.

Let $J_n = \int_1^e(\log x)^n\,dx$. Since $\big[x(\log x)^n\big]_1^e = e$ for $n \ge 1$, $J_n = e - nJ_{n-1}$, with $J_0 = e - 1$: $J_1 = 1$, $J_2 = e - 2$, $J_3 = e - 3(e - 2) = 6 - 2e$.

$I_n = \int_0^{\pi/2}\cos^n x\,dx$. $u = \cos^{n-1}x$, $dv = \cos x\,dx$.

$du = -(n - 1)\cos^{n-2}x\sin x\,dx$, $v = \sin x$: $I_n = \Big[\cos^{n-1}x\sin x\Big]_0^{\pi/2} + (n - 1)\int_0^{\pi/2}\cos^{n-2}x\sin^2 x\,dx$.

The bracket is $0$. With $\sin^2 x = 1 - \cos^2 x$: $I_n = (n - 1)I_{n-2} - (n - 1)I_n$, so $nI_n = (n - 1)I_{n-2}$.

$I_0 = \tfrac{\pi}{2}$, so $I_6 = \tfrac56\cdot\tfrac34\cdot\tfrac12\cdot\tfrac{\pi}{2} = \tfrac{5\pi}{32}$.

Nothing to substitute: the argument of $f''$ is just $x$.

$u = x$, $dv = f''(x)\,dx$, so $du = dx$, $v = f'(x)$.

$\Big[xf'(x)\Big]_0^1 - \int_0^1 f'(x)\,dx = f'(1) - \big(f(1) - f(0)\big)$.

$5 - (4 - 1) = 2$.

$t = \sqrt{x}$, $dx = 2t\,dt$, and $x: 1 \to 9$ is $t: 1 \to 3$: $\int_1^3 2t\,f''(t)\,dt$.

$u = 2t$, $dv = f''(t)\,dt$, so $du = 2\,dt$, $v = f'(t)$.

$\Big[2tf'(t)\Big]_1^3 - 2\int_1^3 f'(t)\,dt = 6f'(3) - 2f'(1) - 2\big(f(3) - f(1)\big)$.

$24 - 2 - 2(7 - 2) = 12$.