Solution. Every part needs rewriting as a sum of powers before the power rule can be applied.

(a). The \(t^{-1}\) term is the excluded case \(n=-1\), so it integrates to a logarithm: \[\int \left (3t^2-t^{-1}\right )dt=\frac {3t^3}{3}-\ln |t|+c=t^3-\ln |t|+c.\]

(b). Split the fraction, which is allowed because the denominator is a single term: \[\frac {2+x}{x^3}=\frac {2}{x^3}+\frac {x}{x^3}=2x^{-3}+x^{-2}.\] \[\int \left (2x^{-3}+x^{-2}+3\right )dx=\frac {2x^{-2}}{-2}+\frac {x^{-1}}{-1}+3x+c =-\frac {1}{x^2}-\frac {1}{x}+3x+c.\]

(d). Expand first: \[\left (x^2+3\right )(x-1)=x^3-x^2+3x-3,\] \[\int \left (x^3-x^2+3x-3\right )dx=\frac {x^4}{4}-\frac {x^3}{3}+\frac {3x^2}{2}-3x+c.\]

(e). Expand the numerator, then divide term by term by \(x^{\frac {1}{2}}\): \[\frac {(2x+1)^2}{\sqrt {x}}=\frac {4x^2+4x+1}{x^{\frac {1}{2}}} =4x^{\frac {3}{2}}+4x^{\frac {1}{2}}+x^{-\frac {1}{2}}.\] \[\int \left (4x^{\frac {3}{2}}+4x^{\frac {1}{2}}+x^{-\frac {1}{2}}\right )dx =\frac {4x^{\frac {5}{2}}}{\frac {5}{2}}+\frac {4x^{\frac {3}{2}}}{\frac {3}{2}} +\frac {x^{\frac {1}{2}}}{\frac {1}{2}}+c =\frac {8}{5}x^{\frac {5}{2}}+\frac {8}{3}x^{\frac {3}{2}}+2\sqrt {x}+c.\]

(f). \[\frac {\sqrt {x}+6x^3}{x}=x^{-\frac {1}{2}}+6x^2,\] so \[\int \left (3+x^{-\frac {1}{2}}+6x^2\right )dx=3x+2\sqrt {x}+2x^3+c.\]

(g). Expand the bracket, then multiply through: \[\sqrt {x}\left (x+6\sqrt {x}+9\right )=x^{\frac {3}{2}}+6x+9x^{\frac {1}{2}},\] \[\int \left (x^{\frac {3}{2}}+6x+9x^{\frac {1}{2}}\right )dx =\frac {2}{5}x^{\frac {5}{2}}+3x^2+6x^{\frac {3}{2}}+c.\]

Solution. (a). \[5e^x-4(-\cos x)+\frac {2x^4}{4}+c=5e^x+4\cos x+\frac {x^4}{2}+c.\]

(b). \[2\left (-\cos x-\sin x+\frac {x^2}{2}\right )+c=-2\cos x-2\sin x+x^2+c.\]

(c). Write \(\frac {2}{x^3}\) as \(2x^{-3}\): \[5e^x+4\sin x-\frac {2x^{-2}}{-2}+c=5e^x+4\sin x+\frac {1}{x^2}+c.\]

(d). The inside is linear, so the answer is the ordinary one divided by the coefficient of \(x\): \[\int \sin (2x+1)\,dx=-\frac {1}{2}\cos (2x+1)+c.\]

(e). \[\int e^{4x-1}dx=\frac {1}{4}e^{4x-1}+c.\]

(f). Expand first — there is no rule for a squared bracket: \[\left (e^{2x}+1\right )^2=e^{4x}+2e^{2x}+1,\] \[\int \left (e^{4x}+2e^{2x}+1\right )dx=\frac {1}{4}e^{4x}+e^{2x}+x+c.\]

Solution. (a). \(u=2x+1\), \(du=2\,dx\): \[\tfrac {1}{2}\int u^5du=\frac {u^6}{12}+c=\frac {(2x+1)^6}{12}+c.\]

(b). \(u=1-2x\), \(du=-2\,dx\): \[3\int u^{-3}\left (-\tfrac {1}{2}\,du\right )=-\tfrac {3}{2}\cdot \frac {u^{-2}}{-2}+c =\frac {3}{4u^2}+c=\frac {3}{4(1-2x)^2}+c.\]

(c). \(u=2x+5\), so \(x=\frac {u-5}{2}\) and \(dx=\frac {du}{2}\): \[\int \frac {u-5}{2}\sqrt {u}\cdot \frac {du}{2}=\tfrac {1}{4}\int \left (u^{\frac {3}{2}} -5u^{\frac {1}{2}}\right )du=\tfrac {1}{4}\left (\tfrac {2}{5}u^{\frac {5}{2}} -\tfrac {10}{3}u^{\frac {3}{2}}\right )+c\] \[=\frac {u^{\frac {3}{2}}}{30}\left (3u-25\right )+c =\frac {(2x+5)^{\frac {3}{2}}(6x-10)}{30}+c=\frac {(2x+5)^{\frac {3}{2}}(3x-5)}{15}+c.\]

(d). \(u=4+x^2\), \(du=2x\,dx\): \[\tfrac {1}{2}\int u^{10}du=\frac {u^{11}}{22}+c=\frac {\left (4+x^2\right )^{11}}{22}+c.\]

(e). The numerator is exactly the derivative of the denominator: \[\int \frac {2x}{x^2+1}dx=\ln \left (x^2+1\right )+c.\] No modulus is needed, since \(x^2+1>0\) always.

(f). Again top is the derivative of bottom, since \(\frac {d}{dx}\left (1+e^x\right )=e^x\): \[\int \frac {e^x}{1+e^x}dx=\ln \left (1+e^x\right )+c.\]

(g). \(\frac {d}{dx}\sin x=\cos x\), so the same pattern: \[\int \frac {\cos x}{\sin x}dx=\ln |\sin x|+c.\]

(h). Here \(\frac {d}{dx}\cos x=-\sin x\), so a minus sign appears: \[u=\cos x,\ du=-\sin x\,dx\quad \implies \quad \int \frac {-du}{u}=-\ln |u|+c =-\ln |\cos x|+c.\] This is \(\int \tan x\,dx\).

Solution. Throughout, \(\int u\,dv=uv-\int v\,du\).

(a). \(u=x\), \(dv=\sin 2x\,dx\), so \(du=dx\) and \(v=-\frac {1}{2}\cos 2x\): \[-\frac {x}{2}\cos 2x+\frac {1}{2}\int \cos 2x\,dx =-\frac {x}{2}\cos 2x+\frac {1}{4}\sin 2x+c.\]

(c). There is no standard integral for \(\ln x\), so take \(u=3\ln x\) and \(dv=dx\), giving \(du=\frac {3}{x}dx\) and \(v=x\): \[3x\ln x-\int x\cdot \frac {3}{x}dx=3x\ln x-3x+c=3x(\ln x-1)+c.\]

(d). Apply the formula twice. First \(u=x^2\), \(dv=e^{-x}dx\), so \(du=2x\,dx\) and \(v=-e^{-x}\): \[-x^2e^{-x}+2\int xe^{-x}dx.\] For the remaining integral, \(u=x\) and \(v=-e^{-x}\): \[\int xe^{-x}dx=-xe^{-x}+\int e^{-x}dx=-xe^{-x}-e^{-x}.\] Combining, \[-x^2e^{-x}+2\left (-xe^{-x}-e^{-x}\right )+c=-e^{-x}\left (x^2+2x+2\right )+c.\]

(e). \(u=x\), \(v=\frac {1}{2}\sin 2x\): \[\frac {x}{2}\sin 2x-\frac {1}{2}\int \sin 2x\,dx=\frac {x}{2}\sin 2x+\frac {1}{4}\cos 2x+c.\]

(f) and (g). These two are done together, because neither factor ever becomes simpler and the integral returns to itself. Write \[I=\int e^x\sin 2x\,dx,\qquad J=\int e^x\cos 2x\,dx.\] Taking \(u=e^x\) and \(dv=\sin 2x\,dx\) in \(I\), so \(v=-\frac {1}{2}\cos 2x\): \[I=-\tfrac {1}{2}e^x\cos 2x+\tfrac {1}{2}\int e^x\cos 2x\,dx =-\tfrac {1}{2}e^x\cos 2x+\tfrac {1}{2}J.\] Similarly, with \(u=e^x\) and \(dv=\cos 2x\,dx\), so \(v=\frac {1}{2}\sin 2x\): \[J=\tfrac {1}{2}e^x\sin 2x-\tfrac {1}{2}\int e^x\sin 2x\,dx =\tfrac {1}{2}e^x\sin 2x-\tfrac {1}{2}I.\] Substituting the second into the first: \[I=-\tfrac {1}{2}e^x\cos 2x+\tfrac {1}{2}\left (\tfrac {1}{2}e^x\sin 2x -\tfrac {1}{2}I\right )=-\tfrac {1}{2}e^x\cos 2x+\tfrac {1}{4}e^x\sin 2x-\tfrac {1}{4}I.\] Collecting the \(I\) terms, \[\tfrac {5}{4}I=\tfrac {1}{4}e^x\left (\sin 2x-2\cos 2x\right )\quad \implies \quad I=\frac {e^x\left (\sin 2x-2\cos 2x\right )}{5}+c.\] Putting that back into the expression for \(J\): \[J=\frac {e^x\left (2\sin 2x+\cos 2x\right )}{5}+c.\]

(h). \(u=x\), \(dv=e^{2x}dx\), so \(v=\frac {1}{2}e^{2x}\): \[\frac {x}{2}e^{2x}-\frac {1}{2}\int e^{2x}dx=\frac {x}{2}e^{2x}-\frac {1}{4}e^{2x}+c =\frac {e^{2x}(2x-1)}{4}+c.\]

Solution. (a). \(3x+5=A(x+2)+B(x+1)\). At \(x=-1\): \(2=A\). At \(x=-2\): \(-1=-B\), so \(B=1\). \[\int \left (\frac {2}{x+1}+\frac {1}{x+2}\right )dx=2\ln |x+1|+\ln |x+2|+c.\]

(b). \(3x-1=A(x-2)+B(2x+1)\). At \(x=2\): \(5=5B\), so \(B=1\). At \(x=-\frac {1}{2}\): \(-\frac {5}{2}=-\frac {5}{2}A\), so \(A=1\). \[\int \left (\frac {1}{2x+1}+\frac {1}{x-2}\right )dx=\tfrac {1}{2}\ln |2x+1|+\ln |x-2|+c.\] The \(\frac {1}{2}\) appears because the derivative of \(2x+1\) is \(2\).

(c). \(2x-6=A(x-1)+B(x+3)\). At \(x=1\): \(-4=4B\), so \(B=-1\). At \(x=-3\): \(-12=-4A\), so \(A=3\). \[\int \left (\frac {3}{x+3}-\frac {1}{x-1}\right )dx=3\ln |x+3|-\ln |x-1|+c.\]

(d). The numerator has degree \(3\) and the denominator degree \(2\), so divide first. With \(x(x+1)=x^2+x\), \[\frac {x^3+2x^2+2}{x^2+x}=x+1+\frac {-x+2}{x^2+x}.\] Now \(-x+2=A(x+1)+Bx\). At \(x=0\): \(2=A\). At \(x=-1\): \(3=-B\), so \(B=-3\). \[\int \left (x+1+\frac {2}{x}-\frac {3}{x+1}\right )dx =\frac {x^2}{2}+x+2\ln |x|-3\ln |x+1|+c.\]

(e). Equal degrees, so divide: \[\frac {x^2}{x^2-4}=1+\frac {4}{x^2-4}=1+\frac {4}{(x-2)(x+2)}.\] Then \(4=A(x+2)+B(x-2)\) gives \(A=1\) at \(x=2\) and \(B=-1\) at \(x=-2\): \[\int \left (1+\frac {1}{x-2}-\frac {1}{x+2}\right )dx=x+\ln |x-2|-\ln |x+2|+c.\]

(f). As printed the denominator reads \((3x+2x)(2-x)^2\), which would collapse to \(5x(2-x)^2\); that is plainly a typo for \((3x+2)(2-x)^2\), and the question is worked in that form. (The part is also labelled “(b)” a second time on the sheet.)

A repeated linear factor needs one term for each power: \[\frac {17-5x}{(3x+2)(2-x)^2}=\frac {A}{3x+2}+\frac {B}{2-x}+\frac {C}{(2-x)^2},\] \[17-5x=A(2-x)^2+B(3x+2)(2-x)+C(3x+2).\] At \(x=2\): \(17-10=7=8C\), so \(C=\frac {7}{8}\). At \(x=-\frac {2}{3}\): \(17+\frac {10}{3}=\frac {61}{3}=A\left (\frac {8}{3}\right )^2 =\frac {64A}{9}\), so \(A=\frac {183}{64}\). Comparing coefficients of \(x^2\): \(0=A-3B\), so \(B=\frac {A}{3}=\frac {61}{64}\).

Now integrate, noting \(\int \frac {dx}{2-x}=-\ln |2-x|\) and \(\int \frac {dx}{(2-x)^2}=\frac {1}{2-x}\): \[\int \left (\frac {183}{64(3x+2)}+\frac {61}{64(2-x)}+\frac {7}{8(2-x)^2}\right )dx\] \[=\frac {183}{64}\cdot \frac {1}{3}\ln |3x+2|-\frac {61}{64}\ln |2-x| +\frac {7}{8}\cdot \frac {1}{2-x}+c\] \[=\frac {61}{64}\ln |3x+2|-\frac {61}{64}\ln |2-x|+\frac {7}{8(2-x)}+c.\]

Solution. (a). \[\int _2^5x^3dx=\left [\frac {x^4}{4}\right ]_2^5=\frac {625}{4}-\frac {16}{4} =\frac {609}{4}=152.25.\]

(b). \[\int _1^33x^{-2}dx=\left [\frac {3x^{-1}}{-1}\right ]_1^3 =\left [-\frac {3}{x}\right ]_1^3=-1-(-3)=2.\]

(c). Divide through first: \[\frac {x^3+2x^2}{x}=x^2+2x,\] \[\int _1^3\left (x^2+2x\right )dx=\left [\frac {x^3}{3}+x^2\right ]_1^3 =\left (9+9\right )-\left (\tfrac {1}{3}+1\right )=18-\tfrac {4}{3}=\frac {50}{3}.\]

Solution.

(a). \(-3x^2+17x-10=-(3x-2)(x-5)\), with roots \(\frac {2}{3}\) and \(5\). Neither lies in \((1,3)\), so there is no crossing, and testing \(x=2\) gives \(-12+34-10=12>0\): the curve is above the axis throughout. \[A=\int _1^3\left (-3x^2+17x-10\right )dx=\left [-x^3+\frac {17x^2}{2}-10x\right ]_1^3\] \[=\left (-27+\tfrac {153}{2}-30\right )-\left (-1+\tfrac {17}{2}-10\right ) =\tfrac {39}{2}-\left (-\tfrac {5}{2}\right )=22.\] \[\therefore \quad A=22 \text { square units}.\]

(b). \(2x^3+7x^2-4x=x(x+4)(2x-1)\), with roots \(-4\), \(0\) and \(\frac {1}{2}\). None lies in \((-3,-1)\), and at \(x=-2\) the value is \(-16+28+8=20>0\): above the axis throughout. \[A=\int _{-3}^{-1}\left (2x^3+7x^2-4x\right )dx =\left [\frac {x^4}{2}+\frac {7x^3}{3}-2x^2\right ]_{-3}^{-1}\] \[=\left (\tfrac {1}{2}-\tfrac {7}{3}-2\right ) -\left (\tfrac {81}{2}-63-18\right )=-\tfrac {23}{6}-\left (-\tfrac {81}{2}\right ) =\frac {-23+243}{6}=\frac {220}{6}=\frac {110}{3}.\] \[\therefore \quad A=\frac {110}{3}=36\tfrac {2}{3} \text { square units}.\]

(c). Factorise: \[-x^4+7x^3-11x^2+5x=-x\left (x^3-7x^2+11x-5\right )=-x(x-5)(x-1)^2.\] The roots are \(0\), \(1\) (repeated) and \(5\). Only \(0\) and \(1\) lie in \([0,4]\), and \(x=0\) is an endpoint.

The root at \(x=1\) is repeated, so the curve touches the axis there without crossing it. Testing either side confirms this: at \(x=0.5\) the value is \(-0.5(-4.5)(0.25)=0.5625>0\), and at \(x=2\) it is \(-2(-3)(1)=6>0\). The curve stays above the axis on the whole interval, so no split is needed. \[A=\int _0^4\left (-x^4+7x^3-11x^2+5x\right )dx =\left [-\frac {x^5}{5}+\frac {7x^4}{4}-\frac {11x^3}{3}+\frac {5x^2}{2}\right ]_0^4\] \[=-\frac {1024}{5}+\frac {1792}{4}-\frac {704}{3}+\frac {80}{2}-0 =-204.8+448-234.667+40=48.533.\] Exactly, \[A=\frac {728}{15}=48\tfrac {8}{15} \text { square units}.\]