# Examples On Integration By Parts Set-3

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## Integration by parts examples

Example – 45

Evaluate the following integrals.

 (a) \begin{align}\int {\sqrt {{x^2} + x + 1} \,\,dx} \end{align} (b) \begin{align}\int {\sqrt {1 - 2x - {x^2}} \,\,dx}\end{align} (c) \begin{align}\int {\sqrt {{x^2} + 3x + 2} \,\,dx}\end{align} (d) \begin{align}\int {x\,\,{e^x}\cos x\,\,dx} \end{align}

Solution: (a) We can write $${x^2} + x + 1$$ as

\begin{align} & {x^2} + x + 1 = \frac{3}{4} + {\left( {x + \frac{1}{2}} \right)^2}\\ &\qquad\qquad\;\;= {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} + {\left( {x + \frac{1}{2}} \right)^2}\end{align}

Thus, this integral is of the standard form (26):

$I = \frac{1}{2}\left\{ {\left( {x + \frac{1}{2}} \right)\sqrt {{x^2} + x + 1} + \frac{3}{4}\ln \left| {\left( {x + \frac{1}{2}} \right) + \sqrt {{x^2} + x + 1} } \right|} \right\} + C$

(b) We can write $$1 - 2x - {x^2}$$ upon rearrangement as

\begin{align} 1 - 2x - {x^2} = 2 - (1 + 2x + {x^2})\\ = {\left( {\sqrt 2 } \right)^2} - {\left( {x + 1} \right)^2}\end{align}

This integral therefore of  the standard form (27) :

$I = \frac{1}{2}\left\{ {(x + 1)\sqrt {1 - 2x - {x^2}} + 2{{\sin }^{ - 1}}\left( {\frac{{x + 1}}{{\sqrt 2 }}} \right)} \right\} + C$

(c) Upon rearrangement, $${x^2} + 3x + 2$$ can be written as

${x^2} + 3x + 2 = {\left( {x + \frac{3}{2}} \right)^2} - \frac{1}{4}\\ \qquad\qquad\qquad\quad = {\left( {x + \frac{3}{2}} \right)^2} - {\left( {\frac{1}{2}} \right)^2}$

Thus, this integral is of the standard form (28):

$I = \frac{1}{2}\left\{ {\left( {x + \frac{3}{2}} \right)\sqrt {{x^2} + 3x + 2} - \frac{1}{4}\ln \left| {\left( {x + \frac{3}{2}} \right) + \sqrt {{x^2} + 3x + 2} } \right.} \right\} + C$

(d) From the standard form (30), we know the integral of $${{e}^{x}}\cos x$$ :

$\int {{e^x}\cos x\,\,dx = \frac{1}{2}{e^x}\left( {\sin x + \cos x} \right)} + C$

Thus, we can apply integration by parts on the expression $$x\,{{e}^{x}}\cos x,taking\;x$$, taking as the first function:

\begin{align}& I = \int {\mathop x\limits_{{\rm{Ist}}} ({e^x}\mathop {\cos x}\limits_{{\rm{IInd}}} )dx} \\ &= \frac{1}{2}x{e^x}(\sin x + \cos x) - \int {1 \cdot \left( {\frac{1}{2}{e^x}(\sin x + \cos x)} \right)dx} \\& = \frac{1}{2}x{e^x}(\sin x + \cos x) - \frac{1}{2}\left\{ {{e^x}\sin x\,\,dx + \int {{e^x}\cos x\,\,dx} } \right\}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\mathop \nearrow \limits_{{\rm{standard form (29)}}} \mathop \nearrow \limits_{{\rm{standard form (30)}}} \\& = \frac{1}{2}x{e^x}(\sin x + \cos x) - \frac{{{e^x}}}{4}\left\{ {(\sin x - \cos x) + (\cos x + \sin x)} \right\} + C\\& = \frac{1}{2}x{e^x}(\sin x + \cos x) - \frac{1}{2}{e^x}\sin x + C\end{align}

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We now consider an integral of the form $$\int {L(x)\sqrt {Q(x)} \,\,dx}$$ where $$L(x)$$ is a linear factor and $$Q(x)$$ is a quadratic factor. As described earlier, we can always find constants $$\alpha \,\,{\rm{and }}\beta$$ such that

$L(x) = \alpha Q'(x) + \beta$

so that

\begin{align}& I = \int {L(x)\sqrt {Q(x)} \,\,dx}\\&= \alpha \int {Q'(x)\sqrt {Q(x)} \,\,dx + \beta \int {\sqrt {Q(x)} \,\,dx} } \\& = \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{I_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;{I_2}\end{align}

$${I_1}$$ can be solved by the substitution $$Q(x) = t$$ so that $$Q'(x)dx = dt$$ and $${I_1}$$ reduces to $$\alpha \int {\sqrt t \,\,dt}$$ whereas $${I_2}$$ is of the standard form (26), (27) or (28) depending on the form of $$Q(x)$$ .

Example – 46

Evaluate the following integrals:

 (a) $$\int {(x + 3)\sqrt {{x^2} + 2x + 3} } \,\,dx$$ (b) \begin{align}\int {\frac{x}{{x - \sqrt {{x^2} - 1} }}} \,\,dx\end{align}

Solution : (a) The derivative of $${x^2} + 2x + 3 \;is \;2x + 3$$ . Thus, we can find $$\alpha \,\,{\rm{and }}\beta$$ such that

\begin{align}&\qquad (x + 3) = \alpha (2x + 3) + \beta\\& \Rightarrow\quad \alpha =\frac{1}{2},\,\,\,\beta =\frac{3}{2} \\& \Rightarrow\quad I = \frac{1}{2}\int {(2x + 3)\sqrt {{x^2} + 2x + 3} } \,\,dx + \frac{3}{2}\int {\sqrt {{x^2} + 2x + 3} } \,\,dx\\ &\qquad\quad = \mathop {\frac{1}{2}\int {\sqrt t } \,\,dt\,\,\,\,\,\,}\limits_{({\text{where }}t{\text{ = }}{x^{\rm{2}}} + 2x + 3)} + \mathop {\frac{3}{2}\int {\sqrt {{{(x + 1)}^2} + 2} } \,\,dx}\limits_{({\text{of the standard form (26)}})} \\&\qquad\quad = \frac{{{t^{3/2}}}}{3} + \frac{3}{4}\left\{ {(x + 1)\sqrt {{x^2} + 2x + 3} + 2\ln \left| {(x + 1) + \sqrt {{x^2} + 2x + 3} } \right|} \right\} + C\\ &\qquad\quad= \frac{{{{({x^2} + 2x + 3)}^{3/2}}}}{3} + \frac{3}{4}\left\{ {(x + 1)\sqrt {{x^2} + 2x + 3} + 2\ln \left| {(x + 1) + \sqrt {{x^2} + 2x + 3} } \right|} \right\} + C\end{align}

(b) We first rationalize the denominator:

\begin{align}& I = \int {\frac{x}{{x - \sqrt {{x^2} - 1} }}\,\,dx}\\& = \int {x(x + \sqrt {{x^2} - 1} )\,\,dx} \\& = \int {{x^2}dx + \mathop {\int {x\sqrt {{x^2} - 1} } \,\,dx}\limits_{{\text{(substitute }}{x^{\text{2}}}{\text{-1 = }}t{\text{)}}} } \\& = \frac{{{x^3}}}{3} + \frac{1}{2}\int {\sqrt t \,\,dt} \\& = \frac{{{x^3}}}{3} + \frac{{{t^{3/2}}}}{3} + C\\& = \frac{{{x^3}}}{3} + \frac{{{{({x^2} - 1)}^{3/2}}}}{3} + C\end{align}

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