Question 4

For each of the following integrals, give an appropriate substitution that will enable the integral to be carried out.
Also give the integral that results from the substitution.
You do not need to evaluate the subsequent integral.

• $${\int \frac{1}{x - \sqrt[3]{x}} \hspace{3pt}dx}$$.
  
  We substitute:
  • $${u = \sqrt[3]{x} = x^{\frac{1}{3}} }$$, so $x = u^3$,
  • $${dx = \frac{dx}{du} du = 3u^2 du}$$.
  So the integral becomes:
  $$\int \frac{1}{x - \sqrt[3]{x}} \hspace{3pt}dx = \int \frac{1}{u^3 - u } \hspace{3pt}3u^2\, \, du$$
  $$= 3\int \frac{u^2}{u(u^2 - 1)} \, \, du = 3\int \frac{u}{(u^2 - 1)} \, \, du.$$
  To finish, we make a second substitution:
  • $${t = u^2 - 1 }$$,
  • $dt = \frac{dt}{du} du = 2u du$,
  So the integral becomes:
  $$3\int \frac{u}{(u^2 - 1)} \, \, du = \frac{3}{2} \int \frac{dt}{t} = \frac{3}{2} \ln(At)$$
  $$= \frac{3}{2} \ln(A(u^2 - 1))$$
  $$= \frac{3}{2} \ln(A( x^{\frac{2}{3}} - 1)).$$

• $${\int_{e}^{e^4} \frac{(\ln(t))^5}{t} \hspace{3pt}dt}$$.
  
  We substitute:
  • $${u = \ln(t)}$$,
  • $${du = \frac{du}{dt}dt = \frac{dt}{t}}$$.
  • When $t = e$, we have $u = \ln(e) = 1$.
  • When $t = e^4$, we have $u = \ln(e^4) = 4$.
  So the integral becomes:
  $$\int_{e}^{e^4} \frac{(\ln(t))^5}{t} \hspace{3pt}dt = \int_{1}^{4}... ...}{6}\right]_1 ^4 = \frac{4^6 - 1}{6} = \frac{4095}{6} = \frac{1365}{2} = 682.5.$$

• $${\int_1^4 \frac{x^3}{(1 + x^2)^{3}} \hspace{3pt}dx}$$.
  
  We substitute:
  • $${u = 1 + x^2}$$, so $x^2 = u - 1$,
  • $${du = \frac{du}{dx}dx = 2x dx }$$.
  • When $x = 1$, we have $u = 1 + 1^2 = 2$.
  • When $x = 4$, we have $u = 1 + 4^2 = 17$.
  So the integrand becomes:
  $$\frac{x^3}{(1 + x^2)^{3}} \hspace{3pt}dx = \frac{x^2(x dx)}{1 + x^2)^3} = \frac{ (u - 1) du}{2u^3}.$$
  Then the integral is:
  $$\int_1^4 \frac{x^3}{(1 + x^2)^{3}} \hspace{3pt}dx = \int_2^{17} \... ...{ (u - 1) du}{2u^3} =\frac{1}{2} \int_2^{17} (\frac{u}{u^3} - \frac{1}{u^3}) du$$
  $$=\frac{1}{2} \int_2^{17} (u^{-2} - u^{-3}) du =\frac{1}{2} \left... ...2}u^{-2}\right]_2^{17} = \frac{1}{4} \left[ \frac{1 - 2u}{u^{2}}\right]_2^{17}$$
  $$= \frac{1}{4} \left(- \frac{33}{289} - (-\frac{3}{4})\right) = \frac{1}{16(289)} (-132 + 867) = \frac{735}{4624} = 0.1589532872.$$