Question
$\int\frac{1}{(7\text{x}-5)^2}+\frac{1}{\sqrt{5\text{x}-4}}\text{dx}$
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Answer
Let $\text{I}=\int\bigg[\frac{1}{(7\text{x}-5)^3}+\frac{1}{\sqrt{5\text{x}-4}}\bigg]\text{dx}.$ Then,
$\text{I}=\int(7\text{x}-5)^{-3}\text{dx}+\int(5\text{x}-4)^{\frac{-1}{2}}\text{dx}$
$=\frac{(7\text{x}-5)^{-2}}{7\times(-2)}+\frac{(5\text{x}-4)^{\frac{1}{2}}}{5\times\frac{1}{2}}+\text{c}$
$=-\frac{(7\text{x}-5)^{6-2}}{14}+\frac{2}{5}\sqrt{(5\text{x}-4)}+\text{c}$
$\text{I}=\frac{-1}{14}(7\text{x}-5)^{-2}+\frac{2}{5}\times\sqrt{5\text{x}-4}+\text{c}.$
$\text{I}=\int(7\text{x}-5)^{-3}\text{dx}+\int(5\text{x}-4)^{\frac{-1}{2}}\text{dx}$
$=\frac{(7\text{x}-5)^{-2}}{7\times(-2)}+\frac{(5\text{x}-4)^{\frac{1}{2}}}{5\times\frac{1}{2}}+\text{c}$
$=-\frac{(7\text{x}-5)^{6-2}}{14}+\frac{2}{5}\sqrt{(5\text{x}-4)}+\text{c}$
$\text{I}=\frac{-1}{14}(7\text{x}-5)^{-2}+\frac{2}{5}\times\sqrt{5\text{x}-4}+\text{c}.$
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