Solution. (a) The bracket is \(x^3-1\) and its derivative is \(3x^2\), which is sitting right there in front. Let \[u=x^3-1\quad \implies \quad \frac {du}{dx}=3x^2\quad \implies \quad du=3x^2\,dx.\] The whole of \(3x^2\,dx\) is replaced by \(du\) in one step: \[\int 3x^2\left (x^3-1\right )^4dx=\int u^4\,du=\frac {u^5}{5}+c =\frac {\left (x^3-1\right )^5}{5}+c.\]

(b) The derivative of the denominator \(x^2+x+5\) is \(2x+1\), which is exactly the numerator. Let \[u=x^2+x+5\quad \implies \quad du=(2x+1)\,dx.\] \[\int \frac {2x+1}{x^2+x+5}\,dx=\int \frac {1}{u}\,du=\ln |u|+c=\ln \left |x^2+x+5\right |+c.\] Whenever the top is the derivative of the bottom, the answer is the logarithm of the bottom. This is worth recognising on sight.

(c) Let \(u=x^2\), so \(du=2x\,dx\): \[\int 2xe^{x^2}dx=\int e^u\,du=e^u+c=e^{x^2}+c.\]

(d) The derivative of \(\cos x\) is \(-\sin x\), so let \[u=\cos x\quad \implies \quad du=-\sin x\,dx\quad \implies \quad \sin x\,dx=-du.\] \[\int \frac {\sin x}{\cos x}\,dx=\int \frac {-du}{u}=-\ln |u|+c=-\ln |\cos x|+c.\] The minus sign comes from the substitution and must not be dropped. This integral is \(\int \tan x\,dx\) in disguise.

(e) Let \(u=x^2+4\), so \(du=2x\,dx\), which means \(x\,dx=\frac {1}{2}du\): \[\int \frac {x}{\sqrt {x^2+4}}\,dx=\frac {1}{2}\int u^{-\frac {1}{2}}du =\frac {1}{2}\cdot \frac {u^{\frac {1}{2}}}{\frac {1}{2}}+c=\sqrt {u}+c=\sqrt {x^2+4}+c.\]

Solution. (a) Take the algebraic factor as \(u\): \[u=x,\quad dv=\sin x\,dx\quad \implies \quad du=dx,\quad v=-\cos x.\] \[\int x\sin x\,dx=uv-\int v\,du=-x\cos x-\int (-\cos x)\,dx=-x\cos x+\int \cos x\,dx\] \[=-x\cos x+\sin x+c.\] The new integral \(\int \cos x\,dx\) is standard, which confirms the choice of \(u\) was the right way round. Had we taken \(u=\sin x\), the new integral would have contained \(\frac {x^2}{2}\cos x\) — worse than the original.

(b) Here \(t\) is the algebraic factor: \[u=2t,\quad dv=e^{3t}dt\quad \implies \quad du=2\,dt,\quad v=\frac {1}{3}e^{3t}.\] \[\int 2te^{3t}dt=2t\cdot \frac {1}{3}e^{3t}-\int \frac {1}{3}e^{3t}\cdot 2\,dt =\frac {2}{3}te^{3t}-\frac {2}{3}\int e^{3t}dt\] \[=\frac {2}{3}te^{3t}-\frac {2}{3}\cdot \frac {1}{3}e^{3t}+c =\frac {2}{3}te^{3t}-\frac {2}{9}e^{3t}+c.\] Note the \(\frac {1}{3}\) appearing each time \(e^{3t}\) is integrated; forgetting it is the usual slip.

(c) The algebraic factor is \(x^2\), which needs differentiating twice before it becomes a constant — so the formula is applied twice.

First application: \(u=x^2\), \(dv=e^x dx\), giving \(du=2x\,dx\) and \(v=e^x\): \[\int x^2e^xdx=x^2e^x-\int 2xe^xdx.\]

Second application, on \(\int 2xe^xdx\): now \(u=2x\), \(dv=e^xdx\), giving \(du=2\,dx\) and \(v=e^x\): \[\int 2xe^xdx=2xe^x-\int 2e^xdx=2xe^x-2e^x+c.\]

Combine: \[\int x^2e^xdx=x^2e^x-\left (2xe^x-2e^x\right )+c=e^x\left (x^2-2x+2\right )+c.\] The bracket must be subtracted as a whole; dropping the brackets flips the sign of the last term.

(d) A logarithm is present, so \(\ln x\) takes the role of \(u\) even though \(x^2\) is algebraic: \[u=\ln x,\quad dv=x^2dx\quad \implies \quad du=\frac {1}{x}dx,\quad v=\frac {x^3}{3}.\] \[\int x^2\ln x\,dx=\frac {x^3}{3}\ln x-\int \frac {x^3}{3}\cdot \frac {1}{x}\,dx =\frac {x^3}{3}\ln x-\frac {1}{3}\int x^2dx\] \[=\frac {x^3}{3}\ln x-\frac {x^3}{9}+c.\] The point of choosing \(u=\ln x\) is visible at the middle step: the \(\frac {1}{x}\) cancels against the \(x^3\) and the logarithm disappears from the remaining integral.

Solution. (a) Factorise the denominator first: \[x^2-x-2=(x-2)(x+1).\] The factors are distinct and linear, so write \[\frac {5x-1}{(x-2)(x+1)}=\frac {A}{x-2}+\frac {B}{x+1}.\] Multiplying through by \((x-2)(x+1)\): \[5x-1=A(x+1)+B(x-2).\] Choose values of \(x\) that kill one bracket at a time: \[x=2:\quad 9=3A\quad \implies \quad A=3,\] \[x=-1:\quad -6=-3B\quad \implies \quad B=2.\] Now integrate each piece: \[\int \left (\frac {3}{x-2}+\frac {2}{x+1}\right )dx=3\ln |x-2|+2\ln |x+1|+c.\]

(b) The factor is repeated, so three terms are needed. It is quicker here to substitute \(u=x-1\), so \(x=u+1\) and \(dx=du\): \[3x+1=3(u+1)+1=3u+4,\] \[\int \frac {3x+1}{(x-1)^3}dx=\int \frac {3u+4}{u^3}\,du =\int \left (3u^{-2}+4u^{-3}\right )du.\] Splitting the fraction is legitimate because the denominator is a single term. Now integrate: \[=\frac {3u^{-1}}{-1}+\frac {4u^{-2}}{-2}+c=-\frac {3}{u}-\frac {2}{u^2}+c =-\frac {3}{x-1}-\frac {2}{(x-1)^2}+c.\] Note that a repeated factor produces powers, not logarithms — only the \(\frac {1}{u}\) term would have given a logarithm, and here there is none.

(c) The denominator has a quadratic factor, so \[\frac {3x^2+4}{x\left (x^2+4\right )}=\frac {A}{x}+\frac {Bx+C}{x^2+4}.\] Multiplying through by \(x\left (x^2+4\right )\): \[3x^2+4=A\left (x^2+4\right )+(Bx+C)x.\] Putting \(x=0\) gives \(4=4A\), so \(A=1\). Comparing coefficients of \(x^2\): \[3=A+B\quad \implies \quad B=2,\] and comparing coefficients of \(x\) gives \(0=C\). Hence \[\int \left (\frac {1}{x}+\frac {2x}{x^2+4}\right )dx.\] The second piece has a numerator that is the derivative of its denominator, so it integrates to a logarithm directly: \[=\ln |x|+\ln \left (x^2+4\right )+c.\] No modulus is needed on the second logarithm, since \(x^2+4\) is positive for every \(x\).

(d) Here the numerator has degree \(3\) and the denominator degree \(2\). Partial fractions require the top to be of lower degree than the bottom, so divide first: \[\frac {x^3+3x^2-2}{x^2+2x}=x+1+\frac {-2x-2}{x^2+2x}.\] Now split the remainder, with \(x^2+2x=x(x+2)\): \[\frac {-2x-2}{x(x+2)}=\frac {A}{x}+\frac {B}{x+2}\quad \implies \quad -2x-2=A(x+2)+Bx.\] \[x=0:\quad -2=2A\quad \implies \quad A=-1,\] \[x=-2:\quad 2=-2B\quad \implies \quad B=-1.\] So \[\int \frac {x^3+3x^2-2}{x^2+2x}\,dx=\int \left (x+1-\frac {1}{x}-\frac {1}{x+2}\right )dx\] \[=\frac {x^2}{2}+x-\ln |x|-\ln |x+2|+c.\]