Question

Find $ \int {(x + 3)\sqrt {3 - 4x - {x^2}} dx} $

Answer 1

Hint: In common words, integration can be defined as bringing together and uniting things. In differential calculus, we have to find the derivative or differential of a given function. But integration is the inverse process of differentiation. In integration, we have to find the function whose differentiation is given. Integrals of this type can be converted into standard form and then solved with the help of some basic formulas of integration.

Complete step-by-step answer:
Let $ I = \int {(x + 3)\sqrt {3 - 4x - {x^2}} } $
The given integral is of the form $ \int {(px + q)\sqrt {a{x^2} + bx + c} dx} $
To simplify this equation, let $ x + 3 = A\dfrac{d}{{dx}}(3 - 4x - {x^2}) + B $
Differentiating the right-hand side, we get
 $
  x + 3 = A( - 4 - 2x) + B \\
   \Rightarrow x + 3 = - 4A - 2Ax + B \\
   \Rightarrow x + 3 = - 4A + B - 2Ax \\
  $
Now on comparing both sides, we get
 $
   - 2A = 1 \\
   \Rightarrow A = - \dfrac{1}{2} \\
  $
And $ - 4A + B = 3 $
Put the value of A in the above equation, we get -
 $
   - 4( - \dfrac{1}{2}) + B = 3 \\
   \Rightarrow 2 + B = 3 \\
   \Rightarrow B = 1 \;
  $
Thus, we can say that $ x + 3 = - \dfrac{1}{2}( - 4 - 2x) + 1 $
Putting this value in the equation given in the question,
 $ I = - \dfrac{1}{2}\int {( - 4 - 2x)\sqrt {3 - 4x - {x^2}} } dx + \int {\sqrt {3 - 4x - {x^2}} } dx $
Now let $ I = {I_1} + {I_2} $ where $ {I_1} = - \dfrac{1}{2}\int {( - 4 - 2x)\sqrt {3 - 4x - {x^2}} dx} $ and $ {I_2} = \sqrt {3 - 4x - {x^2}} dx $
Let us solve $ {I_1} $ first,
Put $ 3 - 4x - {x^2} = t $
On differentiating both sides, we get
 $
   - 4 - 2x = \dfrac{{dt}}{{dx}} \\
   \Rightarrow ( - 4 - 2x)dx = dt \;
  $
Now substitute the value of $ ( - 4 - 2x) $ as $ dt $ in $ {I_1} $
 $
  {I_1} = - \dfrac{1}{2}\int {\sqrt t dt} \\
   \Rightarrow {I_1} = - \dfrac{1}{2}(\dfrac{2}{3}{t^{\dfrac{3}{2}}}) + {C_1} \\
   \Rightarrow {I_1} = - \dfrac{1}{3}{(t)^{\dfrac{3}{2}}} + {C_1} \;
  $
Substitute the value of $ t $ back as $ 3 - 4x - {x^2} $
 $ {I_1} = - \dfrac{1}{3}{(3 - 4x - {x^2})^{\dfrac{3}{2}}} + {C_1} $
Now,
 $
  {I_2} = \sqrt {(3 - 4x - {x^2})} dx \\
  {I_2} = \sqrt { - ({x^2} + 4x - 3)} dx \\
  $
We can rewrite the above equation as,
 $
  {x^2} + 4x - 3 = {x^2} + 4x + 4 - 4 - 3 \\
  {x^2} + 4x - 3 = {x^2} + 4x + 4 - 7 \\
  {x^2} + 4x - 3 = {(x + 2)^2} - {(\sqrt 7 )^2} \;
  $
Putting this value in $ {I_2} $ we have,
 $
  {I_2} = \int {\sqrt { - [{{(x + 2)}^2} - {{(\sqrt 7 )}^2}]} } dx \\
  {I_2} = \int {\sqrt {[{{(\sqrt 7 )}^2} - {{(x + 2)}^2}]} dx} \;
  $
We know that $ \int {\sqrt {{a^2} - {x^2}} dx = } [\dfrac{1}{2}x\sqrt {{a^2} - {x^2}} + \dfrac{{{a^2}}}{2}{\sin ^{ - 1}}(\dfrac{x}{a}) + c] $
Using this formula to solve $ {I_2} $ we get –
 $
  {I_2} = [\dfrac{{x + 2}}{2}\sqrt {7 - {{(x + 2)}^2}} + \dfrac{7}{2}{\sin ^{ - 1}}\dfrac{{x + 2}}{{\sqrt 7 }}] + {C_2} \\
  {I_2} = [\dfrac{{x + 2}}{2}\sqrt {3 - 4x - {x^2}} + \dfrac{7}{2}{\sin ^{ - 1}}\dfrac{{x + 2}}{{\sqrt 7 }}] + {C_2} \;
  $
Putting the value of $ {I_1} $ and $ {I_2} $ in $ I $ , we get –
 $ I = - \dfrac{1}{3}{(3 - 4x - {x^2})^{\dfrac{3}{2}}} + \dfrac{1}{2}(x + 2)\sqrt {3 - 4x - {x^2}} + \dfrac{7}{2}{\sin ^{ - 1}}\dfrac{{x + 2}}{{\sqrt 7 }} + C $
Where $ C = {C_1} + {C_2} $
Thus, $ \int {(x + 2)\sqrt {3 - 4x - {x^2}} dx = - \dfrac{1}{3}{{(3 - 4x - {x^2})}^{\dfrac{3}{2}}} + \dfrac{1}{2}(x + 2)\sqrt {3 - 4x - {x^2}} + \dfrac{7}{2}{{\sin }^{ - 1}}\dfrac{{x + 2}}{{\sqrt 7 }} + C} $

Note: A definite integral is an integral expressed with upper and lower limits while an indefinite integral is expressed without limits. The derivative of a function is unique but integral or anti-derivative of a function can be infinite. Here, C is an arbitrary constant by varying which one can get different values of integral of a function.