Question

Integration x raised to -5/4

Answer 1

Answer:

-4x^-1/4 + c

Step-by-step explanation:

Use the formula,

integration x^n dx = (x^(n+1))/(n+1) .

Answer 2

Answer:

$\displaystyle{\boxed{\red{\sf\:\int\:\left(\:x^{\small{-\:\frac{5}{4}}}\:\right)\:dx\:=\:-\:4\:x^{\small{-\:\frac{1}{4}}}}}}$

Step-by-step-explanation:

We have to integrate

$\displaystyle{\sf\:x^{\small{-\:\frac{5}{4}}}}$

We know that,

Integration by parts is

$\displaystyle{\boxed{\pink{\sf\:\int\:uv'\:=\:uv\:-\:\int\:u'v\:}}}$

Where,

• u is a function
• u' is derivative of u
• v' is another function
• v is integral of v'

Let the given integrand be I.

$\displaystyle{\sf\:I\:=\:\int\:\left(\:x^{\small{-\:\frac{5}{4}}}\:\right)\:dx}$

$\displaystyle{\implies\sf\:I\:=\:\int\:\left(\:x^{\small{-\:\frac{5}{4}}}\:\times\:1\:\right)\:dx}$

Let,

$\displaystyle{\sf\:u\:=\:x^{-\:\frac{5}{4}}}$

$\displaystyle{\sf\:v'\:=\:1}$

Now,

$\displaystyle{\sf\:u'\:=\:\dfrac{d}{dx}\:(\:u\:)}$

$\displaystyle{\implies\sf\:u'\:=\:\dfrac{d}{dx}\:\left(\:x^{\small{-\:\frac{5}{4}}}\:\right)}$

We know that,

$\displaystyle{\boxed{\blue{\sf\:\dfrac{d}{dx}\:(\:x^n\:)\:=\:n\:x^{n\:-\:1}\:}}}$

$\displaystyle{\implies\sf\:u'\:=\:-\:\dfrac{5}{4}\:.\:x^{\small{-\:\frac{5}{4}}\:-\:1}}$

$\displaystyle{\implies\sf\:u'\:=\:-\:\dfrac{5}{4}\:.\:x^{\small{\frac{-\:5\:-\:4}{4}}}}$

$\displaystyle{\implies\sf\:u'\:=\:-\:\dfrac{5}{4}\:.\:x^{\small{-\:\frac{9}{4}}}}$

$\displaystyle{\implies\:\boxed{\purple{\sf\:u'\:=\:-\:\dfrac{5\:x^{\small{-\:\frac{9}{4}}}}{4}\:}}}$

Now,

$\displaystyle{\sf\:v\:=\:\int\:v'}$

$\displaystyle{\implies\sf\:v\:=\:\int\:1}$

We know that, for a constant k,

$\displaystyle{\boxed{\green{\sf\:\int\:k\:dx\:=\:kx\:+\:C\:}}}$

$\displaystyle{\implies\sf\:v\:=\:1\:\times\:x}$

$\displaystyle{\implies\:\boxed{\purple{\sf\:v\:=\:x\:}}}$

Now,

$\displaystyle{\sf\:I\:=\:\int\:\left(\:x^{\small{-\:\frac{5}{4}}}\:\times\:1\:\right)\:dx}$

By using integration by parts,

$\displaystyle{\sf\:\int\:uv'\:=\:uv\:-\:\int\:u'v\:}$

$\displaystyle{\implies\sf\:I\:=\:x^{\small{-\:\frac{5}{4}}}\:\times\:x\:-\:\int\:\left(\:-\:\dfrac{5\:x^{\small{-\:\frac{9}{4}}}}{4}\:\times\:x\:\right)\:dx\:}$

$\displaystyle{\implies\sf\:I\:=\:x^{\small{-\:\frac{5}{4}\:+\:1}}\:-\:\int\:\left(\:-\:\dfrac{5\:x^{\small{-\:\frac{9}{4}\:+\:1}}}{4}\:\right)\:dx}$

$\displaystyle{\implies\sf\:I\:=\:x^{\small{\frac{-\:5\:+\:4}{4}}}\:-\:\int\:\left(\:-\:\dfrac{5\:x^{\small{\frac{-\:9\:+\:4}{4}}}}{4}\:\right)\:dx}$

$\displaystyle{\implies\sf\:I\:=\:x^{\small{-\:\frac{1}{4}}}\:-\:\int\:\left(\:-\:\dfrac{5\:x^{\small{-\:\frac{5}{4}}}}{4}\:\right)\:dx}$

We know that,

$\displaystyle{\boxed{\blue{\sf\:\int\:(\:k\:x\:)\:dx\:=\:k\:\int\:x\:dx\:}}}$

$\displaystyle{\implies\sf\:I\:=\:x^{\small{-\:\frac{1}{4}}}\:-\:\left(\:-\:\dfrac{5}{4}\:\right)\:\underbrace{\sf\:\int\:\left(\:x^{\small{-\:\frac{5}{4}}}\:\right)\:dx}_{\large{\sf\:I}}}$

$\displaystyle{\implies\sf\:I\:=\:x^{\small{-\:\frac{1}{4}}}\:+\:\dfrac{5}{4}\:I}$

$\displaystyle{\implies\sf\:I\:-\:\dfrac{5\:I}{4}\:=\:x^{\small{-\:\frac{1}{4}}}}$

$\displaystyle{\implies\sf\:\dfrac{4I\:-\:5I}{4}\:=\:x^{\small{-\:\frac{1}{4}}}}$

$\displaystyle{\implies\sf\:\dfrac{-\:I}{4}\:=\:x^{\small{-\:\frac{1}{4}}}}$

$\displaystyle{\implies\sf\:-\:I\:=\:4\:x^{\small{-\:\frac{1}{4}}}}$

$\displaystyle{\implies\sf\:I\:=\:-\:4\:x^{\small{-\:\frac{1}{4}}}}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:\int\:\left(\:x^{\small{-\:\frac{5}{4}}}\:\right)\:dx\:=\:-\:4\:x^{\small{-\:\frac{1}{4}}}}}}}$