Question

integrate the following:
(2t -4)^-4 dt = please give detailed answer​

Answer 1

$\huge\mathcal{\fcolorbox{cyan}{black}{\pink{ANSWER}}}$

First we will expand it !

$(2t - 4)^4 \\ = {(2t)}^{4} + \: ^4C_1 \: {(2t)}^{4 - 1} \times ( - 4)^{1} + \: ^4C_2 \: {(2t)}^{4 - 2} \times ( - 4)^{2} + \: ^4C_3\: {(2t)}^{4 - 3} \times ( - 4)^{3} + {( - 4)}^{4} \\ = 16 {t}^{4} - 128 {t}^{3} + 384 {t}^{2} - 512 {t} + 256$

Now,

$\displaystyle \int \: {(2t - 4)}^{4} \: dt \\ = \int \: (16 {t}^{4} - 128 {t}^{3} + 384 {t}^{2} - 512 {t} + 256) \: dt \\ = \int \: 16 {t}^{4} \: dt - \int 128 {t}^{3} \: dt + \int \: 384 {t}^{2} \: dt - \int \: 512 {t} \: dt+ \int \: 256 \: dt \\ = 16 \times \frac{ {t}^{5} }{5} - 128 \times \frac{ {t}^{4} }{4} + 384 \times \frac{ {t}^{3} }{3} - 512 \times \frac{ {t}^{2} }{2} + 256t \:+\:c\\ = \frac{16 {t}^{5} }{5} - 32 {t}^{4} + 128 {t}^{3} - 256 {t}^{2} + 256t\:+\:c$