Question #26a54

Rewrite using fractional and negative exponents:

#=int(4x^(2/3)-8x^(-9/5))dx#

Use the rule #intx^ndx=x^(n+1)/(n+1)+C#, where #n!=-1#:

#=4(x^(5/3)/(5/3))-8(x^(-4/5)/(-4/5))+C#

#=12/5x^(5/3)+10x^(-4/5)+C#

We can write this in the original format if we want:

#=12/5root3(x^5)+10/root5(x^4)+C#

Use the sum rule to break the big integral into littler ones:
#int4root(3)(x)^2dx-int8/(root(5)(x)^9)dx#

Now take out the constant terms (the things without an #x#):
#4introot(3)(x)^2dx-8int1/(root(5)(x)^9)dx#

Think back to algebra and your exponent rules. How do you express #root(a)(x)^b# using exponents? The rule is:
#root(a)(x)^b=x^(b/a)#

For our problem, this means
#root(3)(x)^2=x^(2/3)#
#1/(root(5)(x)^9)=1/x^(9/5)=x^(-9/5)#

(Also recall that #1/x^a = x^(-a)#).

Now we have:
#4intx^(2/3)dx-8intx^(-9/5)dx#

These integrals are easily solved using the reverse power rule. Remember that to differentiate, we brought the power to the front and then reduced the power by one; so the derivative to #4x^3#, for example, is
#3*4x^(3-1)=12x^2#

Since integration is the opposite of differentiation, we do the opposite here: increase the power by one and divide by the new power. In general,
#intx^adx=1/(a+1)(x^(a+1))+C#

That means
#4intx^(2/3)dx=4(1/(2/3+1))(x^(2/3+1))=4*3/5*x^(5/3)=12/5x^(5/3)#
#8intx^(-9/5)dx=8(1/(-9/5+1))(x^(-9/5+1))=8*-5/4*x^(-5/4)=-10x^(-5/4)#

Putting it all together, we have:
#4intx^(2/3)dx-8intx^(-9/5)dx#
#=12/5x^(5/3)-(-10x^(-5/4))#

For a final answer of #12/5x^(5/3)+10x^(-5/4)+C#. Don't forget the constant of integration #C#!