Question

Find the integrals (primitives):
$\rm \displaystyle\int \bigg[\frac{1}{(7x-5)^{3}} -\frac{1}{\sqrt{5x-4}} \bigg] dx$

Answer 1

To find the integral of function,we have to apply linearity and power rule of Integration

$\int \bigg[\frac{1}{(7x-5)^{3}} -\frac{1}{\sqrt{5x-4}} \bigg] dx \\ \\ \int \: (7x-5)^{ - 3} dx- \int \: {(5x - 4)}^{ \frac{ - 1}{2} } dx$
as we know that in integration the coefficient of x had to placed in denominator

$\int \: {x}^{n} dx = \frac{ {x}^{n + 1} }{n + 1}$
So,

$= \frac{ {(7x - 5)}^{ - 3 + 1} }{ 7(- 3 + 1)} - \frac{ {(5x - 4)}^{( \frac{ - 1}{2} + 1) } }{ 5(\frac{ - 1}{2} + 1) } + C \\ \\ = \frac{ {(7x - 5)}^{ - 2} }{ 7(- 2)} - \frac{ {(5x - 4)}^{( \frac{ 1}{2}) } }{ 5(\frac{ 1}{2} ) } + C\\ \\ = \frac{ - 1}{14( {7x - 5)}^{2} } - \frac{2 \sqrt{5x - 4} }{5} + C$
Thus

$\int \bigg[\frac{1}{(7x-5)^{3}} -\frac{1}{\sqrt{5x-4}} \bigg] dx \\ \\ = \frac{ - 1}{14( {7x - 5)}^{2} } - \frac{2 \sqrt{5x - 4} }{5} + C$
Hope it helps you.