Integral (2x-3)⁸ dx.

1.	Integral (2x-3)⁸ dx.
Answer» $\sf \: Let \: I \: = \displaystyle\int \tt \: {(<klux>2X</klux> - 3)}^{8} dx$ As we know that, By using substitution method, we get. 2x - 3 = t. DIFFERENTIATE w.r.t x, we get. ⇒ 2 dx = dt. ⇒ dx = dt/2. So, GIVEN INTEGRAL can be rewritten as $\rm :\longmapsto\:I \: = \displaystyle\int \tt \: {t}^{8} \: \dfrac{dt}{2}$ $\rm :\longmapsto\: = \dfrac{ {t}^{8 + 1} }{(8 + 1) \times 2} + c$ $\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf \: \because \: \displaystyle\int \sf \: {x}^{n}dx = \dfrac{ {x}^{n + 1} }{n + 1} + c}$ $\rm :\longmapsto\: = \tt \: \dfrac{ {(2x - 3)}^{9} }{18} + c$ ─━─━─━─━─━─━─━─━─━─━─━─━─ Additional INFORMATION :- $\boxed{ \bf{ \:\displaystyle\int \sf \: \dfrac{1}{x} dx = log(x) + c}}$ $\boxed{ \bf{ \:\displaystyle\int \sf \: sinxdx = - cosx + c}}$ $\boxed{ \bf{ \:\displaystyle\int \sf \: cosxdx = sinx + c}}$ $\boxed{ \bf{ \:\displaystyle\int \sf \: {sec}^{2}x = tanx + c}}$ $\boxed{ \bf{ \:\displaystyle\int \sf \: {cosec}^{2}xdx = - cotx}}$ $\boxed{ \bf{ \:\displaystyle\int \sf \: {e}^{x}dx = {e}^{x} + c}}$

Answer»

$\sf \: Let \: I \: = \displaystyle\int \tt \: {(<klux>2X</klux> - 3)}^{8} dx$