# Examples On Integration By Substitution Set-5

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Example - 19

Evaluate the following integrals

 (a) \begin{align} \int {\frac{1}{{\sqrt {4x - {x^2} - 3} }}dx} \end{align} (b)\begin{align} \int {\frac{1}{{{x^2} + 2x + 2}}dx} \end{align} (c) \begin{align} \int {\frac{1}{{(x + 2)\sqrt {{x^2} + 4x + 3} }}dx} \end{align} (d) \begin{align} \int {\frac{1}{{{x^2} + 6x + 8}}\,} dx \end{align} (e) \begin{align} \int {\frac{1}{{\sqrt {4{x^2} + 12x + 10} }}dx} \end{align} (f)\begin{align} \int {\frac{1}{{\sqrt {25{x^2} + 20x + 3} }}dx} \end{align}

Solution: If you observe the forms of the expressions to be integrated carefully, you will realise that each part corresponds to a part in the previous example. Hence we’ll use the results obtained in the previous part directly:

(a) $I = \int {\frac{1}{{\sqrt {4x - {x^2} - 3} }}dx}$

\begin{align}&\,\,\,\, = \int {\frac{1}{{\sqrt {1 - {{(x - 2)}^2}} }}dx} \\&\,\,\,\, = {\sin ^{ - 1}}(x - 2) + C\end{align}

(b) $I = \int {\frac{1}{{{x^2} + 2x + 2}}dx}$

\begin{align}&\,\,\,\, = \int {\frac{1}{{{{(x + 1)}^2} + 1}}dx} \\&\,\,\,\, = {\tan ^{ - 1}}(x + 1) + C\end{align}

(c) $I = \int {\frac{1}{{(x + 2)\sqrt {{x^2} + 4x + 3} }}} dx$

\begin{align}&\,\,\,\, = \int {\frac{1}{{(x + 2)\sqrt {{{(x + 2)}^2} - 1} }}} dx\\&\,\,\,\, = {\sec ^{ - 1}}(x + 2) + C\end{align}

(d) $I = \int {\frac{1}{{{x^2} + 6x + 8}}dx}$

\begin{align}&\,\,\,\, = \int {\frac{1}{{{{(x + 3)}^2} - 1}}dx} \\&\,\,\,\, = \frac{1}{2}\ln \left| {\frac{{x + 2}}{{x + 4}}} \right| + C\end{align}

(e) $I = \int {\frac{1}{{\sqrt {4{x^2} + 12x + 10} }}dx}$

\begin{align}&\qquad\qquad \qquad\;= \int {\frac{1}{{\sqrt {{{(2x + 3)}^2} + 1} }}dx} \\&\qquad\qquad\qquad\;= \frac{1}{2}\ln \left| {(2x + 3) + \sqrt {4{x^2} + 12x + 10} } \right| + C\end{align}

(f) $I = \int {\frac{1}{{\sqrt {25{x^2} + 20x + 3} }}dx}$

\begin{align}&\qquad\qquad\qquad\;\;= \int {\frac{1}{{\sqrt {{{(5x + 2)}^2} - 1} }}dx} \\& \qquad\qquad\qquad\;\; = \frac{1}{5}\ln \left| {(5x + 2) + \sqrt {25{x^2} + 20x + 3} } \right| + C\end{align}

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