Question

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

Answer 1

Hint: The given integration is in the form of $\int{\left( px+q \right)}\sqrt{a{{x}^{2}}+bx+c}$. in such type of integration, we have to assume that $px+q=A.\dfrac{d}{dx}\left( a{{x}^{2}}+bx+c \right)+B$, here $A\text{ and }B$are constant which is to be found. Here we have $p=1,q=3,a=-1,b=-4,c=3$. After this transformation we get the integrand which is easily integrable. After simplification we get sum of two integration in which one can be find by use of substitution method and other is based on standard form.

Complete step by step answer:
We have to integrate $\int{(x+3)\sqrt{3-4x-{{x}^{2}}}}dx$let us suppose this is $I$, so we can write
$I=\int{(x+3)\sqrt{3-4x-{{x}^{2}}}}dx$
Now we assume that
$x+3=A.\dfrac{d}{dx}\left( 3-4x-{{x}^{2}} \right)+B$
As we now, that $\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}},\dfrac{dc}{dx}=0$, where $c$is a constant, so we can write further
$\begin{align}
  & x+3=A(-4-2x)+B \\
 & \Rightarrow x+3=-4A-2Ax+B \\
 & \Rightarrow x+3=\left( -4A+B \right)+\left( -2Ax \right) \\
\end{align}$
Now in order to find the value of $A\text{ and }B$, we have to compare, so on comparing, we have
$\begin{align}
  & -2A=1 \\
 & \Rightarrow A=-\dfrac{1}{2} \\
\end{align}$
And
$-4A+B=3$
Putting the value of $\Rightarrow A=-\dfrac{1}{2}$, we get
 $\begin{align}
  & -4\left( \dfrac{-1}{2} \right)+B=3 \\
 & \Rightarrow B=3-2 \\
 & \Rightarrow B=1 \\
\end{align}$
So, we can write
$x+3=\left( \dfrac{-1}{2} \right)\left( -4-2x \right)+1$
Now we can write the given integration as
$\begin{align}
  & I=\int{\left\{ -\dfrac{1}{2}(-4-2x)+1 \right\}\sqrt{3-4x-{{x}^{2}}}}dx \\
 & I=\int{\left\{ -\dfrac{1}{2}(-4-2x) \right\}\sqrt{3-4x-{{x}^{2}}}}dx+\int{\sqrt{3-4x-{{x}^{2}}}}dx \\
\end{align}$
Here we get the given integration as the sum of two integration, so, we can write
$I={{I}_{1}}+{{I}_{2}}$
Here ${{I}_{1}}=\int{\left\{ -\dfrac{1}{2}(-4-2x) \right\}\sqrt{3-4x-{{x}^{2}}}}dx$
And ${{I}_{2}}=\int{\sqrt{3-4x-{{x}^{2}}}}dx$
Now we calculate both integrands separately, first we calculate
${{I}_{1}}=\int{\left\{ -\dfrac{1}{2}(-4-2x) \right\}\sqrt{3-4x-{{x}^{2}}}}dx$
Let $3-4x-{{x}^{2}}=t$
Differentiating both sides, we can write
$\left( -4-2x \right)dx=dt$
Now substituting the above, we can write
${{I}_{1}}=\dfrac{-1}{2}\int{\sqrt{t}}dt$
As we know that $\int{{{x}^{n}}}dx=\dfrac{{{x}^{n+1}}}{n+1}+c$
We can write
$\begin{align}
  & {{I}_{1}}=\dfrac{-1}{2}\left( \dfrac{{{t}^{\dfrac{1}{2}+1}}}{\dfrac{1}{2}+1} \right) \\
 & {{I}_{1}}=\left( \dfrac{-1}{2} \right)\left( {{t}^{\dfrac{3}{2}}} \right)\left( \dfrac{2}{3} \right) \\
 & {{I}_{1}}=\left( \dfrac{-1}{3} \right)\left( {{t}^{\dfrac{3}{2}}} \right) \\
\end{align}$
Now we substitute the assumed value of $t$we can write
${{I}_{1}}=\left( \dfrac{-1}{3} \right)\left( {{\left( 3-4x-{{x}^{2}} \right)}^{\dfrac{3}{2}}} \right)+{{c}_{1}}-------------(1)$
Now we have to evaluate ${{I}_{2}}$, we can write
${{I}_{2}}=\int{\sqrt{3-4x-{{x}^{2}}}}dx$
Here we make the terms in square root as the difference of two square, so, we can write
$\begin{align}
  & {{I}_{2}}=\int{\sqrt{7-\left( 4x+{{x}^{2}}+4 \right)}}dx \\
 & \Rightarrow {{I}_{2}}=\int{\sqrt{{{\left( \sqrt{7} \right)}^{2}}-{{\left( x+2 \right)}^{2}}}}dx \\
\end{align}$
Now we put here
$\begin{align}
  & z=x+2 \\
 & dz=dx \\
\end{align}$
Hence, we can write
${{I}_{2}}=\int{\sqrt{{{\left( \sqrt{7} \right)}^{2}}-{{\left( z \right)}^{2}}}}dx$
As we know that
$\int{\sqrt{{{a}^{2}}-{{x}^{2}}}}=\dfrac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}+\dfrac{{{a}^{2}}}{2}{{\sin }^{-1}}\dfrac{x}{a}+C$
Using the above formula, we can write
\[\begin{align}
  & {{I}_{2}}=\int{\sqrt{{{\left( \sqrt{7} \right)}^{2}}-{{\left( z \right)}^{2}}}}dx \\
 & \Rightarrow {{I}_{2}}=\dfrac{z}{2}\sqrt{{{\left( \sqrt{7} \right)}^{2}}-{{z}^{2}}}+\dfrac{{{\left( \sqrt{7} \right)}^{2}}}{2}{{\sin }^{-1}}\dfrac{z}{\sqrt{7}}+{{c}_{2}} \\
\end{align}\]
Now putting the value of $z=x+2$, we can write
\[\begin{align}
  & {{I}_{2}}=\int{\sqrt{{{\left( \sqrt{7} \right)}^{2}}-{{\left( z \right)}^{2}}}}dx \\
 & \Rightarrow {{I}_{2}}=\dfrac{x+2}{2}\sqrt{{{\left( \sqrt{7} \right)}^{2}}-{{\left( x+2 \right)}^{2}}}+\dfrac{{{\left( \sqrt{7} \right)}^{2}}}{2}{{\sin }^{-1}}\dfrac{x+2}{\sqrt{7}}+{{c}_{2}} \\
 & \Rightarrow {{I}_{2}}=\dfrac{x+2}{2}\sqrt{3-4x-{{x}^{2}}}+\dfrac{7}{2}{{\sin }^{-1}}\dfrac{x+2}{\sqrt{7}}+{{c}_{2}}--------(2) \\
\end{align}\]
Now as we have
$I={{I}_{1}}+{{I}_{2}}$
So, adding (1) and (2) we can write
\[\begin{align}
  & I=-\dfrac{1}{3}{{\left( 3-4x-{{x}^{2}} \right)}^{\dfrac{3}{2}}}+\dfrac{x+2}{2}\sqrt{3-4x-{{x}^{2}}}+\dfrac{7}{2}{{\sin }^{-1}}\dfrac{x+2}{\sqrt{7}}+{{c}_{1}}+{{c}_{2}} \\
 & I=-\dfrac{1}{3}{{\left( 3-4x-{{x}^{2}} \right)}^{\dfrac{3}{2}}}+\dfrac{x+2}{2}\sqrt{3-4x-{{x}^{2}}}+\dfrac{7}{2}{{\sin }^{-1}}\dfrac{x+2}{\sqrt{7}}+K \\
\end{align}\]
Here K is the combined integration constant.

Note:
If we have to evaluate the integral of type $\int{f\{\phi (x)\}{\phi }'(x)dx}$ then we have to put $\phi (x)=t\text{ and }{\phi }'(x)dx=dt$. With this substitution, the integrand becomes easily integrable.