 $\displaystyle \int{{\frac{{7x+3}}{{7x-4}}dx}}$

Separate the single fraction into two fractions.  The objective is to remove variable x from the numerator (See explanation at the bottom of post). $\displaystyle \int{{\frac{{7x-4+7}}{{7x-4}}dx}}$ $\displaystyle \int{{\frac{{7x-4}}{{7x-4}}+\frac{7}{{7x-4}}dx}}$

Simplify and integrate the expression. $\displaystyle \int{{1+\frac{7}{{7x-4}}dx}}$ $\displaystyle \int{{1dx+\int{{\frac{7}{{7x-4}}}}dx}}$ $\displaystyle \int{{1dx+7\int{{\frac{1}{{7x-4}}}}dx}}$ $\displaystyle x+\frac{{7\ln (7x-4)}}{7}+c$ $\displaystyle x+\ln (7x-4)+c$

Unlike differentiation, there is no quotient rule for integration, therefore we have to either remove the x variable from the numerator or denominator.

In the solution above we separate the fraction so that the x varable in the numerator is removes. Once it is two separate fraction we integrate the expression. Do note that for the fraction on the right $\displaystyle \int{{\frac{1}{{(ax+b)}}=\frac{{\ln (ax+b)}}{a}}}+c$

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