Integration: past paper questions & worked solutions

$10x^4-\dfrac{3}{2x^3}-7=10x^4-\dfrac32x^{-3}-7$.

$\int10x^4\,dx=2x^5$, $\int-\dfrac32x^{-3}\,dx=\dfrac34x^{-2}$, and $\int-7\,dx=-7x$.

$2x^5+\dfrac{3}{4x^2}-7x+c$.

$8x^3-6\sqrt{x}-\dfrac{2}{5x^3}=8x^3-6x^{1/2}-\dfrac25x^{-3}$.

$\int8x^3\,dx=2x^4$, $\int-6x^{1/2}\,dx=-4x^{3/2}$, and $\int-\dfrac25x^{-3}\,dx=\dfrac15x^{-2}$.

$2x^4-4x^{3/2}+\dfrac{1}{5x^2}+c$.

$\int\left(10x^5+6x^3-\dfrac{3}{x^2}\right)dx=\int(10x^5+6x^3-3x^{-2})dx$.

$\int 10x^5dx=\dfrac{10x^6}{6}=\dfrac53x^6$, $\int 6x^3dx=\dfrac{6x^4}{4}=\dfrac32x^4$, and $\int -3x^{-2}dx=3x^{-1}$.

Therefore the integral is $\dfrac53x^6+\dfrac32x^4+\dfrac3x+c$.

$\dfrac{8x^3}{5}-\dfrac{2}{3x^4}-1=\dfrac85x^3-\dfrac23x^{-4}-1$.

$\int\dfrac85x^3dx=\dfrac25x^4$, $\int-\dfrac23x^{-4}dx=\dfrac29x^{-3}$, and $\int-1dx=-x$.

Therefore the integral is $\dfrac{2x^4}{5}+\dfrac{2}{9x^3}-x+c$.

$\dfrac{8x^3}{3}-\dfrac{1}{2\sqrt{x}}-5=\dfrac83x^3-\dfrac12x^{-1/2}-5$.

$\int\dfrac83x^3\,dx=\dfrac23x^4$, $\int-\dfrac12x^{-1/2}\,dx=-x^{1/2}$, and $\int-5\,dx=-5x$.

Therefore $\int\left(\dfrac{8x^3}{3}-\dfrac{1}{2\sqrt{x}}-5\right)dx=\dfrac23x^4-\sqrt{x}-5x+c$.

$\left(3x+\frac{2}{x}\right)^2=9x^2+12+\frac{4}{x^2}=9x^2+12+4x^{-2}$.

$\displaystyle \int\left(9x^2+12+4x^{-2}\right)\,\mathrm{d}x=3x^3+12x-4x^{-1}+C$.

Since $x^{-1}=\dfrac{1}{x}$, the integral is $3x^3+12x-\dfrac{4}{x}+C$.

$(x-3)^2=x^2-6x+9$.

$(x^2-6x+9)(2x+5)=2x^3+5x^2-12x^2-30x+18x+45$.

$(x-3)^2(2x+5)=2x^3-7x^2-12x+45$.

$\int(2x^3-7x^2-12x+45)dx=\dfrac12x^4-\dfrac73x^3-6x^2+45x+c$.

$(x+3)^2=x^2+6x+9$.

$\dfrac{(x+3)^2}{3\sqrt{x}}=\dfrac{x^2+6x+9}{3x^{1/2}}=\dfrac13x^{3/2}+2x^{1/2}+3x^{-1/2}$.

$\int\left(\dfrac13x^{3/2}+2x^{1/2}+3x^{-1/2}\right)dx=\dfrac13\cdot\dfrac25x^{5/2}+2\cdot\dfrac23x^{3/2}+3\cdot2x^{1/2}+c$.

$\dfrac{2}{15}x^{5/2}+\dfrac43x^{3/2}+6x^{1/2}+c$.

$(2x-5)(3x+2)=6x^2-11x-10$.

$(6x^2-11x-10)(2x+5)=12x^3+8x^2-75x-50$.

$\int(12x^3+8x^2-75x-50)\,dx=3x^4+\dfrac83x^3-\dfrac{75}{2}x^2-50x+c$.

$\dfrac{4x^5+3}{2x^2}=2x^3+\dfrac32x^{-2}$.

$\int2x^3dx=\dfrac{x^4}{2}$.

$\int\dfrac32x^{-2}dx=-\dfrac32x^{-1}$.

The integral is $\dfrac{x^4}{2}-\dfrac{3}{2x}+c$.

$12x^3+\dfrac{1}{6\sqrt{x}}-\dfrac{3}{2x^4}=12x^3+\dfrac16x^{-1/2}-\dfrac32x^{-4}$.

$\int 12x^3dx=3x^4$, $\int\dfrac16x^{-1/2}dx=\dfrac13x^{1/2}$, and $\int-\dfrac32x^{-4}dx=\dfrac12x^{-3}$.

Therefore the integral is $3x^4+\dfrac{\sqrt{x}}{3}+\dfrac{1}{2x^3}+c$.

$y=4x^3+\dfrac{3}{2}x^{-1/2}-8$.

$\dfrac{dy}{dx}=12x^2+\dfrac32\left(-\dfrac12\right)x^{-3/2}$.

$\dfrac{dy}{dx}=12x^2-\dfrac34x^{-3/2}=12x^2-\dfrac{3}{4x^{3/2}}$.

$\int y\,dx=\int\left(4x^3+\dfrac32x^{-1/2}-8\right)dx$.

$\int y\,dx=x^4+3x^{1/2}-8x+c=x^4+3\sqrt{x}-8x+c$.

$(3\sqrt{x}+2)(x-5)=3x\sqrt{x}-15\sqrt{x}+2x-10$.

$\dfrac{(3\sqrt{x}+2)(x-5)}{4\sqrt{x}}=\dfrac{3x}{4}-\dfrac{15}{4}+\dfrac12x^{1/2}-\dfrac52x^{-1/2}$.

$\int\left(\dfrac{3x}{4}-\dfrac{15}{4}+\dfrac12x^{1/2}-\dfrac52x^{-1/2}\right)dx=\dfrac{3x^2}{8}-\dfrac{15x}{4}+\dfrac13x^{3/2}-5x^{1/2}+c$.

So the integral is $\dfrac{3x^2}{8}-\dfrac{15x}{4}+\dfrac{x\sqrt{x}}{3}-5\sqrt{x}+c$.

The normal is $y=-\dfrac1{24}x+5$. At $x=3$, $y=-\dfrac3{24}+5=\dfrac{39}{8}$.

The normal gradient is $-\dfrac1{24}$, so the tangent gradient is $24$.

$f'(3)=4(3)^2+3k+3=39+3k$.

$39+3k=24$, so $k=-5$.

With $k=-5$, $f'(x)=4x^2-5x+3$, so $f(x)=\dfrac43x^3-\dfrac52x^2+3x+c$.

$\dfrac{39}{8}=\dfrac43(27)-\dfrac52(9)+3(3)+c=\dfrac{45}{2}+c$.

$c=\dfrac{39}{8}-\dfrac{45}{2}=-\dfrac{141}{8}$.

$(2x+1)(x+3)=2x^2+7x+3$, so $(x-5)(2x^2+7x+3)=2x^3-3x^2-32x-15$.

Comparing with $ax^3+bx^2-32x-15$ gives $a=2$ and $b=-3$.

$\dfrac{(x-5)(2x+1)(x+3)}{5\sqrt{x}}=\dfrac15(2x^{5/2}-3x^{3/2}-32x^{1/2}-15x^{-1/2})$.

$\int\dfrac15(2x^{5/2})dx=\dfrac{4}{35}x^{7/2}$ and $\int\dfrac15(-3x^{3/2})dx=-\dfrac{6}{25}x^{5/2}$.

$\int\dfrac15(-32x^{1/2})dx=-\dfrac{64}{15}x^{3/2}$ and $\int\dfrac15(-15x^{-1/2})dx=-6x^{1/2}$, then add $c$.

$f'(x)=ax-12x^{1/3}$, so $f''(x)=a-4x^{-2/3}$.

Since $27^{2/3}=9$, $0=a-\dfrac{4}{9}$, so $a=\dfrac49$.

With $a=\dfrac49$, $f'(x)=\dfrac49x-12x^{1/3}$. Therefore $f(x)=\dfrac29x^2-9x^{4/3}+c$.

$-8=\dfrac29-9+c$, so $c=-8+9-\dfrac29=\dfrac79$.

$f(x)=\dfrac29x^2-9x^{4/3}+\dfrac79$.

The tangent has equation $y=5x-3\sqrt2$, so the gradient at $x=\sqrt2$ is $5$.

$f'(\sqrt2)=2\sqrt2+\dfrac8{(\sqrt2)^2}+k=2\sqrt2+4+k=5$, so $k=1-2\sqrt2$.

At $x=\sqrt2$, the tangent gives $y=5\sqrt2-3\sqrt2=2\sqrt2$.

The tangent gradient is $5$, so the normal gradient is $-\dfrac15$.

$y-2\sqrt2=-\dfrac15(x-\sqrt2)$, hence $y=-\dfrac15x+\dfrac{11\sqrt2}{5}$.

$f(x)=\int\left(2x+\dfrac8{x^2}+k\right)dx=x^2-\dfrac8x+kx+c$.

$f(x)=x^2-\dfrac8x+(1-2\sqrt2)x+c$.

$2\sqrt2=2-\dfrac8{\sqrt2}+(1-2\sqrt2)\sqrt2+c=-2-3\sqrt2+c$.

$c=2+5\sqrt2$, so $f(x)=x^2-\dfrac8x+(1-2\sqrt2)x+2+5\sqrt2$.

$f'(x)=4x^{3/2}+kx^{-2}$, so $f''(x)=6x^{1/2}-2kx^{-3}$.

At $P$, $x=2$ and $f''(2)=0$, so $6\sqrt2-2k(2^{-3})=0$.

$6\sqrt2-\dfrac{k}{4}=0$, hence $k=24\sqrt2$.

$f(x)=\int\left(4x^{3/2}+24\sqrt2x^{-2}\right)dx=\dfrac85x^{5/2}-24\sqrt2x^{-1}+c$.

$P(2,8\sqrt2)$ lies on $C$, so $8\sqrt2=\dfrac85(2^{5/2})-\dfrac{24\sqrt2}{2}+c$.

$8\sqrt2=\dfrac{32\sqrt2}{5}-12\sqrt2+c$, so $c=\dfrac{68\sqrt2}{5}$.

Since $P$ has x-coordinate 4 and lies on the tangent $y=3x+4$, $P=(4,16)$.

The tangent gradient is 3, so the normal gradient is $-\dfrac13$. Hence the normal is $y-16=-\dfrac13(x-4)$.

$f''(x)=4x+x^{-1/2}$, so $f'(x)=2x^2+2\sqrt{x}+A$.

At $x=4$, $f'(4)=3$. Thus $2(16)+2(2)+A=3$, so $A=-33$.

$f(x)=\dfrac23x^3+\dfrac43x^{3/2}-33x+B$.

$16=\dfrac23(64)+\dfrac43(8)-33(4)+B=\dfrac{160}{3}-132+B$.

$B=\dfrac{284}{3}$, so $f(x)=\dfrac23x^3+\dfrac43x^{3/2}-33x+\dfrac{284}{3}$.

$f'(x)=\dfrac{32}{3x^2}+3-2\sqrt[3]{x}$. At $x=8$, $f'(8)=\dfrac{32}{3(64)}+3-2(2)=\dfrac16-1=-\dfrac56$.

Using $P(8,2)$, $y-2=-\dfrac56(x-8)$, hence $y=-\dfrac56x+\dfrac{26}{3}$.

$f'(x)=\dfrac{32}{3}x^{-2}+3-2x^{1/3}$, so $f(x)=-\dfrac{32}{3x}+3x-\dfrac32x^{4/3}+c$.

$2=-\dfrac{32}{24}+24-\dfrac32(16)+c=-\dfrac43+c$, so $c=\dfrac{10}{3}$.

$f(x)=-\dfrac{32}{3x}+3x-\dfrac32x^{4/3}+\dfrac{10}{3}$.

$f(x)=\dfrac{(x+5)^2}{\sqrt{x}}=\dfrac{x^2+10x+25}{x^{1/2}}=x^{3/2}+10x^{1/2}+25x^{-1/2}$.

$\int f(x)\,dx=\dfrac25x^{5/2}+\dfrac{20}{3}x^{3/2}+50x^{1/2}+c$.

$f'(x)=\dfrac32x^{1/2}+5x^{-1/2}-\dfrac{25}{2}x^{-3/2}$.

$\dfrac32x^{1/2}+5x^{-1/2}-\dfrac{25}{2}x^{-3/2}=0$.

Multiplying by $2x^{3/2}$ gives $3x^2+10x-25=0$.

$3x^2+10x-25=(3x-5)(x+5)$, so $x=\dfrac53$ or $x=-5$. Since $x>0$, $x=\dfrac53$.

$f'(x)=12x^{-1/2}+\dfrac{x}{3}-4$, so $f(x)=24\sqrt{x}+\dfrac{x^2}{6}-4x+c$.

$8=24\sqrt9+\dfrac{9^2}{6}-4(9)+c=72+\dfrac{81}{6}-36+c$.

$72+\dfrac{81}{6}-36=\dfrac{99}{2}$, so $c=8-\dfrac{99}{2}=-\dfrac{83}{2}$.

$f'(9)=\dfrac{12}{3}+\dfrac93-4=3$.

The normal gradient is the negative reciprocal of 3, so it is $-\dfrac13$.

The normal through $(9,8)$ is $y-8=-\dfrac13(x-9)$, so $y=-\dfrac13x+11$. It crosses the y-axis at $(0,11)$.

$f'(x)=2x^{-1/2}+Ax^{-2}+3$, so $f''(x)=-x^{-3/2}-2Ax^{-3}$.

$-4^{-3/2}-2A(4^{-3})=0$, so $-\dfrac18-\dfrac{A}{32}=0$ and $A=-4$.

$f'(x)=\dfrac2{\sqrt{x}}-\dfrac4{x^2}+3$.

$f(x)=4\sqrt{x}+\dfrac4x+3x+c$.

$f(12)=8\sqrt3+\dfrac13+36+c=8\sqrt3$, so $c=-\dfrac{109}{3}$.

At $x=4$, $f'(4)=7$. Since $f'(x)=\dfrac{2x^2+ax+b}{4\sqrt{x}}$, we get $\dfrac{32+4a+b}{8}=7$.

$32+4a+b=56$, so $4a+b=24$.

Together with $a+b=-9$, subtracting gives $3a=33$, so $a=11$ and $b=-20$.

$f'(x)=\dfrac{2x^2+11x-20}{4\sqrt{x}}=\dfrac12x^{3/2}+\dfrac{11}{4}x^{1/2}-5x^{-1/2}$.

$f(x)=\dfrac15x^{5/2}+\dfrac{11}{6}x^{3/2}-10x^{1/2}+c$.

$-5=\dfrac15(4^{5/2})+\dfrac{11}{6}(4^{3/2})-10(4^{1/2})+c=\dfrac{16}{15}+c$.

$c=-\dfrac{91}{15}$, so $f(x)=\dfrac15x^{5/2}+\dfrac{11}{6}x^{3/2}-10x^{1/2}-\dfrac{91}{15}$.

$y=f(x-3)$ translates the curve 3 units to the right, so $P(4,-5)$ becomes $Q(7,-5)$.

$(2x-1)(3x+2)=6x^2+x-2$.

$f'(x)=6x-\dfrac{6x^2+x-2}{2\sqrt{x}}=6x-3x^{3/2}-\dfrac12x^{1/2}+x^{-1/2}$.

$f'(4)=24-3(8)-\dfrac12(2)+\dfrac12=-\dfrac12$.

The normal gradient is the negative reciprocal of $-\dfrac12$, so it is $2$.

Through $P(4,12)$, $y-12=2(x-4)$, hence $y=2x+4$.

$f(x)=3x^2-\dfrac65x^{5/2}-\dfrac13x^{3/2}+2x^{1/2}+c$.

$12=3(4)^2-\dfrac65(4^{5/2})-\dfrac13(4^{3/2})+2(4^{1/2})+c$.

$12=48-\dfrac{192}{5}-\dfrac83+4+c=\dfrac{164}{15}+c$, so $c=\dfrac{16}{15}$.

$f'(x)=\dfrac{4x^2+10-7x^{1/2}}{4x^{1/2}}=x^{3/2}+\dfrac52x^{-1/2}-\dfrac74$.

At $x=4$, $f'(4)=4^{3/2}+\dfrac52(4^{-1/2})-\dfrac74=8+\dfrac54-\dfrac74=\dfrac{15}{2}$.

The normal gradient is the negative reciprocal, so $m_n=-\dfrac{2}{15}$.

Using $P(4,-1)$, $y+1=-\dfrac{2}{15}(x-4)$.

$15y+15=-2x+8$, hence $2x+15y+7=0$.

$f(x)=\int\left(x^{3/2}+\dfrac52x^{-1/2}-\dfrac74\right)dx=\dfrac25x^{5/2}+5x^{1/2}-\dfrac74x+c$.

$-1=\dfrac25(4^{5/2})+5(4^{1/2})-\dfrac74(4)+c=\dfrac{64}{5}+10-7+c$, so $c=-\dfrac{84}{5}$.

For $y=\dfrac14x^3-8x^{-1/2}$, $\dfrac{dy}{dx}=\dfrac34x^2+4x^{-3/2}$.

At $x=4$, $\dfrac{dy}{dx}=\dfrac34(16)+4(4^{-3/2})=12+\dfrac12=\dfrac{25}{2}$.

Using $P(4,12)$, $y-12=\dfrac{25}{2}(x-4)$.

Multiplying by 2 gives $2y-24=25x-100$, hence $25x-2y-76=0$.

$f'(x)=\dfrac14x^3-8x^{-1/2}$ gives $f(x)=\dfrac{x^4}{16}-16\sqrt{x}+c$.

$12=\dfrac{4^4}{16}-16\sqrt4+c=16-32+c$, so $c=28$.

At $P\left(4,\frac{32}{3}\right)$, $f'(4)=5$, so the tangent has gradient $5$.

$y-\frac{32}{3}=5(x-4)$. Hence $y=5x-\frac{28}{3}$.

$f''(x)=4x^{-1/2}-3$, so $f'(x)=8x^{1/2}-3x+C$.

$5=8(2)-12+C=4+C$, so $C=1$. Therefore $f'(x)=8\sqrt{x}-3x+1$.

$f(x)=\frac{16}{3}x^{3/2}-\frac32x^2+x+D$.

Substitute $x=4$, $f(4)=\frac{32}{3}$: $\frac{32}{3}=\frac{16}{3}(8)-\frac32(16)+4+D=\frac{68}{3}+D$, so $D=-12$.

The tangent gradient at $P$ is $f'\!\left(\frac14\right)$. Using $$f'(x)=10+\frac{5\sqrt{x}-4}{2x^2},$$ we get $$f'\!\left(\frac14\right)=10+\frac{5\left(\frac12\right)-4}{2\left(\frac14\right)^2} =10+\frac{-\frac32}{\frac18} =-2.$$ Therefore $k=-2$.

The tangent is $y=-2x+2$. At $x=\frac14$, $$y=-2\left(\frac14\right)+2=\frac32,$$ so $P=\left(\frac14,\frac32\right)$.

The tangent gradient is $-2$, so the normal gradient is $\frac12$. Through $P$, $$y-\frac32=\frac12\left(x-\frac14\right).$$ Hence $$y=\frac12x+\frac{11}{8}.$$

First split the fractional term: $$f'(x)=10+\frac52x^{-3/2}-2x^{-2}.$$ Integrating term by term gives $$f(x)=10x-5x^{-1/2}+2x^{-1}+c =10x-\frac5{\sqrt{x}}+\frac2x+c.$$

Since $P=\left(\frac14,\frac32\right)$ lies on the curve, $$\frac32=10\left(\frac14\right)-5\left(\frac14\right)^{-1/2}+2\left(\frac14\right)^{-1}+c.$$ Thus $$\frac32=\frac52-10+8+c=\frac12+c,$$ so $c=1$.

The tangent at $P$ has equation $y=10x+c$, so the gradient at $P$ is $10$.

$f'(4)=3\sqrt4+k(4)^2=6+16k=10$, so $16k=4$ and $k=\dfrac14$.

Substitute $P(4,12)$ into $y=10x+c$: $12=40+c$, so $c=-28$.

With $k=\dfrac14$, $f'(x)=3x^{1/2}+\dfrac14x^2$.

$f''(x)=\dfrac32x^{-1/2}+\dfrac12x$.

$f''(4)=\dfrac32\cdot\dfrac12+2=\dfrac34+2=\dfrac{11}{4}$.

$f(x)=\int\left(3x^{1/2}+\dfrac14x^2\right)dx=2x^{3/2}+\dfrac1{12}x^3+d$.

$12=2(4^{3/2})+\dfrac1{12}(4^3)+d=16+\dfrac{16}{3}+d$.

$d=-\dfrac{28}{3}$, so $f(x)=2x^{3/2}+\dfrac1{12}x^3-\dfrac{28}{3}$.

$f'(x)=\dfrac{(x+3)^2}{x\sqrt{x}}=\dfrac{x^2+6x+9}{x^{3/2}}=x^{1/2}+6x^{-1/2}+9x^{-3/2}$.

At $x=4$, $f'(4)=2+3+\dfrac98=\dfrac{49}{8}$.

Using $P(4,20)$, $y-20=\dfrac{49}{8}(x-4)$. Multiplying out gives $49x-8y-36=0$.

$f(x)=\int\left(x^{1/2}+6x^{-1/2}+9x^{-3/2}\right)dx=\dfrac23x^{3/2}+12x^{1/2}-18x^{-1/2}+c$.

$20=\dfrac23(8)+12(2)-\dfrac{18}{2}+c=\dfrac{61}{3}+c$, so $c=-\dfrac13$.

$f(x)=\dfrac23x\sqrt{x}+12\sqrt{x}-\dfrac{18}{\sqrt{x}}-\dfrac13$.

$\dfrac{(3x+2)^2}{4\sqrt{x}}=\dfrac{9x^2+12x+4}{4x^{1/2}}=\dfrac94x^{3/2}+3x^{1/2}+x^{-1/2}$.

$\int\left(\dfrac94x^{3/2}+3x^{1/2}+x^{-1/2}\right)dx=\dfrac9{10}x^{5/2}+2x^{3/2}+2x^{1/2}+c$.

Factoring out $\sqrt{x}$ gives $\dfrac{\sqrt{x}(9x^2+20x+20)}{10}+c$.

Since $f'(x)=x^2+ax+b$ and the gradient at $P(3,-2)$ is $2$, $f'(3)=2$. Hence $9+3a+b=2$, so $3a+b=-7$.

$f(x)=\dfrac{x^3}{3}+\dfrac{a}{2}x^2+bx+c$. The y-intercept is $-8$, so $c=-8$.

$-2=\dfrac{27}{3}+\dfrac{9a}{2}+3b-8$, so $\dfrac{9a}{2}+3b=-3$.

From $3a+b=-7$, $b=-7-3a$. Substitute into $\dfrac{9a}{2}+3b=-3$ to get $a=-4$, and then $b=5$.

Therefore $f(x)=\dfrac{x^3}{3}-2x^2+5x-8$.

At $P$, $f'(4)=-4$, so the tangent gradient is $-4$. The normal gradient is therefore $\frac14$.

Using $P(4,-50)$: $y+50=\frac14(x-4)$, so $y=\frac14x-51$.

$f''(x)=6x^{-3/2}+x$. Therefore $f'(x)=-12x^{-1/2}+\frac12x^2+k$.

$-4=-\frac{12}{2}+\frac12(16)+k=-6+8+k=2+k$, so $k=-6$.

$f(x)=-24x^{1/2}+\frac{x^3}{6}-6x+d$.

Substitute $P(4,-50)$: $-50=-24(2)+\frac{64}{6}-24+d=-72+\frac{32}{3}+d=-\frac{184}{3}+d$. Hence $d=\frac{34}{3}$.

$f'(\sqrt3)=4(\sqrt3)^2+\dfrac6{(\sqrt3)^2}-4=12+2-4=10$, which matches the tangent gradient.

Set $4x^2+\dfrac6{x^2}-4=10$.

Multiplying by $x^2$ gives $4x^4-14x^2+6=0$, so $2x^4-7x^2+3=0$.

$2x^4-7x^2+3=(2x^2-1)(x^2-3)=0$.

$x=\sqrt3$ or $x=\dfrac1{\sqrt2}=\dfrac{\sqrt2}{2}$. The other possible x-coordinate is $\dfrac{\sqrt2}{2}$.

Given $x_P=\sqrt3$, the tangent gives $y_P=10\sqrt3-6\sqrt3=4\sqrt3$.

The normal gradient is $-\dfrac1{10}$, so $y-4\sqrt3=-\dfrac1{10}(x-\sqrt3)$.

$f(x)=\int\left(4x^2+\dfrac6{x^2}-4\right)dx=\dfrac43x^3-\dfrac6x-4x+c$.

$4\sqrt3=\dfrac43(\sqrt3)^3-\dfrac6{\sqrt3}-4\sqrt3+c=4\sqrt3-2\sqrt3-4\sqrt3+c$.

$c=6\sqrt3$, so $f(x)=\dfrac43x^3-\dfrac6x-4x+6\sqrt3$.