Answer
\[ = 3\]
Work Step by Step
\[\begin{gathered}
\int_{ - 3}^1 {\frac{{dx}}{{\,{{\left( {2x + 6} \right)}^{\frac{2}{3}}}}}} \hfill \\
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integration\,\,of\,\,indefinite\,\,integral \hfill \\
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\int_{}^{} {\frac{{dx}}{{\,{{\left( {2x + 6} \right)}^{\frac{2}{3}}}}}} = \int_{}^{} {\,{{\left( {2x + 6} \right)}^{ - \frac{2}{3}}}dx} \hfill \\
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\operatorname{int} egrate\,\,use\,\,the\,\,power\,\,rule \hfill \\
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= \frac{1}{2}\frac{1}{{\,\left( {\frac{1}{3}} \right)}}\,{\left( {2x + 6} \right)^{\frac{1}{3}}} + C \hfill \\
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use\,\,the\,\,Definition\,\,of\,\,improper\,\,integral\, \hfill \\
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\int_a^b {f\,\left( x \right)} \,dx\,\, = \,\,\mathop {\lim }\limits_{c \to {a^ + }} \int_c^b {f\,\left( x \right)} \,dx \hfill \\
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provided\,\,the\,\,\,limit\,\,exists \hfill \\
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\int_{ - 3}^1 {\frac{{dx}}{{\,{{\left( {2x + 6} \right)}^{\frac{2}{3}}}}}} \,\, = \mathop {\lim }\limits_{c \to - {3^ + }} \int_c^1 {\frac{1}{{\,{{\left( {2x + 6} \right)}^{\frac{2}{3}}}}}\,dx} \hfill \\
then \hfill \\
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= \mathop {\lim }\limits_{c \to - {3^ + }} \,\left( {\frac{3}{2}\,{{\left( {2x + 6} \right)}^{\frac{1}{3}}}} \right)_c^1 \hfill \\
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use\,\,the\,\,ftc \hfill \\
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= \frac{3}{2}\,\left( {2 - \mathop {\lim }\limits_{c \to - {3^ + }} \,{{\left( {2c + 6} \right)}^{\frac{1}{3}}}} \right) \hfill \\
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evaluate\,\,the\,\,limit \hfill \\
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= 3 \hfill \\
\end{gathered} \]
