Find
[∫36(3−4x)2 dx [\int \frac{36}{(3 - 4x)^2} \, dx [∫(3−4x)236dx] giving your answer in simplest form.
Express 4x+71−2x\frac{4x + 7}{1 - 2x}1−2x4x+7 in the form
A+B1−2x where A and B are constants to be found. A + \frac{B}{1 - 2x} \text{ where } A \text{ and } B \text{ are constants to be found.} A+1−2xB where A and B are constants to be found.Hence find, using algebraic integration, the exact value of
[∫−404x+71−2x dx [\int_{-4}^{0} \frac{4x + 7}{1 - 2x} \, dx [∫−401−2x4x+7dx] giving your answer in the form alnb−ca\ln b - calnb−c, where a,b,a, b,a,b, and ccc are integers.
