Question

integral (2x-3)⁸ dx. ​

Answer 1

$\sf \: Let \: I \: = \displaystyle\int \tt \: {(2x - 3)}^{8} dx$

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

$\rm :\longmapsto\:I \: = \displaystyle\int \tt \: {t}^{8} \: \dfrac{dt}{2}$

$\rm :\longmapsto\: = \dfrac{ {t}^{8 + 1} }{(8 + 1) \times 2} + c$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf \: \because \: \displaystyle\int \sf \: {x}^{n}dx = \dfrac{ {x}^{n + 1} }{n + 1} + c}$

$\rm :\longmapsto\: = \tt \: \dfrac{ {(2x - 3)}^{9} }{18} + c$

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Additional Information :-

$\boxed{ \bf{ \:\displaystyle\int \sf \: \dfrac{1}{x} dx = log(x) + c}}$

$\boxed{ \bf{ \:\displaystyle\int \sf \: sinxdx = - cosx + c}}$

$\boxed{ \bf{ \:\displaystyle\int \sf \: cosxdx = sinx + c}}$

$\boxed{ \bf{ \:\displaystyle\int \sf \: {sec}^{2}x = tanx + c}}$

$\boxed{ \bf{ \:\displaystyle\int \sf \: {cosec}^{2}xdx = - cotx}}$

$\boxed{ \bf{ \:\displaystyle\int \sf \: {e}^{x}dx = {e}^{x} + c}}$