Question

integrate (7x+8)^5/2 dx
(that 5/2 is in power)​

Answer 1

$\large\underline{\sf{Solution-}}$

$\displaystyle \: \rm :\longmapsto\: \int {\bigg(7x + 8 \bigg) }^{\dfrac{5}{2} } dx$

Here we use method of Substitution.

$\rm :\longmapsto\:Put \: 7x + 8 = y$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}(7x + 8) = \dfrac{d}{dx}y$

$\rm :\longmapsto\:\dfrac{d}{dx}7x + \dfrac{d}{dx}8 = \dfrac{dy}{dx}$

$\rm :\longmapsto\:7 = \dfrac{dy}{dx}$

$\rm :\longmapsto\:dx = \dfrac{dy}{7}$

Thus,

The given integral can be rewritten as,

$\displaystyle \: \rm :\longmapsto\: \int {\bigg(7x + 8 \bigg) }^{\dfrac{5}{2} } dx$

$\displaystyle \: \rm \: = \: \: \int \: {\bigg( y\bigg) }^{\dfrac{5}{2} } \dfrac{dy}{7}$

$\displaystyle \: \rm \: = \: \:\dfrac{1}{7} \int \: {\bigg( y\bigg) }^{\dfrac{5}{2} }dy$

$\rm \: = \: \: \dfrac{1}{7}\dfrac{ {\bigg( y\bigg) }^{\dfrac{5}{2} + 1 } }{\dfrac{5}{2} + 1 } + c$

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

$\rm \: = \: \: \dfrac{1}{7}\dfrac{ {\bigg( y\bigg) }^{\dfrac{5 + 2}{2}} }{\dfrac{5 + 2}{2}} + c$

$\rm \: = \: \: \dfrac{1}{7}\dfrac{ {\bigg( y\bigg) }^{\dfrac{7}{2}} }{\dfrac{7}{2}} + c$

$\rm \: = \: \: \dfrac{2}{49} {\bigg( 7x + 8\bigg) }^{\dfrac{7}{2} } + c$

Additional Information :-

$\green{\boxed{ \bf \: \int \: kdx = kx+ c}}$

$\green{\boxed{ \bf \: \int \: sinx \: dx = - cosx+ c}}$

$\green{\boxed{ \bf \: \int \: cosx \: dx = sinx+ c}}$

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

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

$\green{\boxed{ \bf \: \int \:secx \: tanx dx = secx+ c}}$

$\green{\boxed{ \bf \: \int \:cosecx \: cotx dx = - \: cosecx+ c}}$

$\green{\boxed{ \bf \: \int \: cotx \: dx \: = \: log(sinx) + c}}$

$\green{\boxed{ \bf \: \int \: tanx \: dx \: = \: log(secx) + c}}$

$\green{\boxed{ \bf \: \int \: secx \: dx \: = \: log(secx + tanx) + c}}$