How do you integrate #int (z^2+1)/sqrt(z)dz#?

2 Answers
Gió
Oct 10, 2015

I found: #2(z^(5/2)/5+sqrt(z))+c#

Explanation:

Let us set #z=t^2# so:
#dz=2tdt#
and substituting:
#=int(t^4+1)/cancel(t)(2cancel(t))dt=#
#2int(t^4+1)dt=2(t^5/5+t)+c=#
back to #z#:
#=2(z^(5/2)/5+sqrt(z))+c#

Tony B
Oct 10, 2015

#2/5 z^(1/2) ( z^(4/2) +5) + C => 2/5 sqrt(z) (z^2+5)+C#

Explanation:

Separate and simplify the original expression into two parts then integrate each part. Sum the final integrations:

Write #(z^2+1)/(sqrt(z)) " " # as #" "(z^2)/(z^(1/2)) + 1/(z^(1/2))#

Simplifying the indices giving:

#z^(3/2) + z^(-1/2)#

Now integrate so we have:

#int(z^(3/2)).dz + int(z^(-1/2)).dz#

Giving:

#[ (z^(3/2 +2/2))/(5/2) + C_1 ] + [ (z^(-1/2 + 2/2))/(1/2) + C_2]#

Let #C_1 + C_2 = C_3# giving:

#2/5 x^(5/2) + 2z^(1/2) + C_3#

But #2/5 z^(1/2) times 5 = 2z^(1/2)#

Factoring out # 2/5 z^(1/2)# gives:

#2/5 sqrt(z) (z^2+5)+C_3#

My copy of Maple used as a check of final solution

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