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Question
Solution
\[\int\frac{x}{\sqrt{x + 4}}dx\]
\[ = \int\left( \frac{x + 4 - 4}{\sqrt{x + 4}} \right)dx\]
\[ = \int\left( \sqrt{x + 4} - \frac{4}{\sqrt{x + 4}} \right)dx\]
\[ = \int \left( x + 4 \right)^\frac{1}{2} dx - 4\int \left( x + 4 \right)^{- \frac{1}{2}} dx\]
\[ = \frac{\left( x + 4 \right)^\frac{1}{2} + 1}{\frac{1}{2} + 1} - 4\frac{\left[ x + 4 \right]^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} + C\]
\[ = \frac{2}{3} \left( x + 4 \right)^\frac{3}{2} - 8 \left( x + 4 \right)^\frac{1}{2} + C\]
\[ = \left( x + 4 \right)^\frac{1}{2} \left[ \frac{2}{3}\left( x + 4 \right) - 8 \right] + C\]
\[ = \left( x + 4 \right)^\frac{1}{2} \left[ \frac{2x + 8 - 24}{3} \right] + C\]
\[ = \left( x + 4 \right)^\frac{1}{2} \left[ \frac{2x - 16}{3} \right] + C\]
\[ = \frac{2}{3}\left( x - 8 \right)\sqrt{x + 4} + C\]
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