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• 7 Advanced Integration & Application Ap Calculus Frq

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We now consider more irrational algebraic expressions. The forms that are generally encountered are listed below along with the recommended substitutions:

(a) (begin{align}int {frac{1}{{{L_1}left( x right)sqrt {{L_2}left( x right)} }}} ,dx qquad :;;;{text{Substitute}},{L_2}left( x right) = {t^2}end{align})

(b) (begin{align}int {frac{1}{{Lleft( x right)sqrt {Qleft( x right)} }}} ,dx qquad :;;;;{text{Substitute}},Lleft( x right) = frac{1}{t}end{align})

(c) (begin{align}int {frac{1}{{Qleft( x right)sqrt {Lleft( x right)} }}} ,dx qquad :;;;{text{Substitute}},Lleft( x right) = {t^2}end{align})

(d) (begin{align}int {frac{1}{{{Q_1}left( x right)sqrt {{Q_2}left( x right)} }}} ,dx qquad :;;;;{text{Substitute}},frac{{{Q_2}left( x right)}}{{{Q_1}left( x right)}} = {t^2}left( begin{gathered}{text{If the expressions}}{Q_1}left( x right){text{ and }}{Q_2}left( x right){text{are}},{text{purely quadratic}},{text{i}}{text{.e}},{text{the coefficients}}{text{of }}x,{text{are both are zero}},{text{the}}{text{substitution}};x = frac{1}{t}{text{would also do }}end{gathered} right)end{align})

Some examples will make the use of these substitutions more clear

Example – 56

Evaluate the following integrals:

(a) (begin{align}int {frac{1}{{left( {x - 1} right)sqrt {x + 2} }}} ,dx end{align})

(b) (begin{align}int {frac{x}{{left( {x - 3} right)sqrt {x + 1} }}} ,dxend{align})

(c) (begin{align}int {frac{1}{{left( {x + 1} right)sqrt {{x^2} + x + 1} }}} ,dx end{align})

(d) (begin{align}int {frac{x}{{left( {{x^2} + 2x + 2} right)sqrt {x + 1} }}} ,dx end{align})

Solution: Refer to the substitutions mentioned in the preceeding discussion:

(a) Substitute

[begin{align}& qquad x+2={{t}^{2}} & Rightarrow quad dx=2,tdt & Rightarrowquad I=int{frac{1}{left( x-1 right)sqrt{x+2}}},dx &qquadquad =int{frac{1}{left( {{t}^{2}}-3 right)cdot t}}cdot ,2,tdt &qquadquad =2int{frac{1}{{{t}^{2}}-3}},dt &qquadquad =frac{1}{sqrt{3}}ln left| frac{t-sqrt{3}}{t+sqrt{3}} right|+C &qquadquad =frac{1}{sqrt{3}}ln left| frac{sqrt{x+2}-sqrt{3}}{sqrt{x+2}+sqrt{3}} right|+C end{align}]

(b) Substitute

[begin{align} &qquad x + 1 = {t^2} & Rightarrowquad dx=2tdt & Rightarrow quad I=int{frac{x}{left( x-3 right)sqrt{x+1}}}dx & qquadquad =int{frac{left( {{t}^{2}}-1 right)}{left( {{t}^{2}}-4 right)cdot t}}2tdt & qquadquad=2int{frac{{{t}^{2}}-1}{{{t}^{2}}-4}}dt & qquadquad=2int{left{ 1+frac{3}{{{t}^{2}}-4} right}}dt & qquadquad=2t+frac{3}{2}ln left| frac{t-2}{t+2} right|+C & qquadquad =2sqrt{x+1}+frac{3}{2}ln left| frac{sqrt{x+1}-2}{sqrt{x+1}+2} right|+C end{align}]

(c) Substitute

[begin{align}&qquad x + 1 = frac{1}{t}& Rightarrow quad dx=-frac{1}{{{t}^{2}}}dt & Rightarrow quad I=int{frac{1}{left( x+1 right)sqrt{{{x}^{2}}+x+1}}}dx &qquadquad=int{frac{t}{sqrt{{{left( frac{1}{t}-1 right)}^{2}}+left( frac{1}{t}-1 right)+1}}}cdot left( -frac{1}{{{t}^{2}}} right)dt &qquadquad=-int{frac{1}{sqrt{{{t}^{2}}-t+1}}}dt &qquadquad=-int{frac{1}{sqrt{{{left( t-frac{1}{2} right)}^{2}}+frac{3}{4}}}}dt &qquadquad=-ln left| left( t-frac{1}{2} right)+sqrt{{{t}^{2}}-t+1} right|+C & qquadquad =-ln left| frac{1}{x+1}-frac{1}{2}+frac{sqrt{{{x}^{2}}+x+1}}{x+1} right|+C end{align}]

(d) Substitute

[begin{align} &qquadquad x+1={{t}^{2}} &Rightarrow quad dx=2tdt &Rightarrowquad I=int{frac{x}{left( {{x}^{2}}+2x+2 right)sqrt{x+1}}}dx &qquadquad =int{frac{{{t}^{2}}-1}{left( {{t}^{4}}+1 right)cdot t}}cdot 2tdt & qquadquad=2int{frac{{{t}^{2}}-1}{{{t}^{4}}+1}}dt end{align}]

7 Advanced Integration & Application Ap Calculus Frq

We’ve already seen how to evaluate integrals of this type