How do you integrate A and B?

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2 Answers
Apr 3, 2018

Given: #int x^3e^(4x^4) dx#

Let #u = 4x^4#, then #du = 16x^3 dx# or #x^3dx = 1/16du#:

#int x^3e^(4x^4) dx = 1/16int e^u du#

#int x^3e^(4x^4) dx = 1/16 e^u + C#

Reverse the substitution:

#int x^3e^(4x^4) dx = 1/16 e^(4x^4) + C#

Given: #int 1/(xsqrtx) + e^(5x) dx#

Substitute #1/(xsqrtx) = x^(-3/2)#:

#int 1/(xsqrtx) + e^(5x) dx = int x^(-3/2) + e^(5x) dx#

Separate into two integrals:

#int 1/(xsqrtx) + e^(5x) dx = int x^(-3/2)dx + int e^(5x) dx#

The first integral fits the power rule and the second integral is well known:

#int 1/(xsqrtx) + e^(5x) dx = -2x^(-1/2) + 1/5e^(5x) + C#

Apr 3, 2018

# (a) : 1/16e^(4x^4)+C_(a) #.

# (b) : 1/5e^(5x)-2/sqrtx+C_(b)#.

Explanation:

Part (a) :

Let, #I_(a)=intx^3e^(4x^4)dx#.

Note that, #d/dx(x^4)=4x^3#. So, let us subst.

#y=4x^4 :. dy=4*4x^3dx=16x^3dx#.

#:. I_(a)=intx^3e^(4x^4)dx=1/16inte^(4x^4)*(16x^3)dx#,

#=1/16inte^ydy#.

# rArr I_(a)=1/16e^(4x^4)+C_(a)#.

Part (b) :

Suppose that, #I_(b)=int(1/(xsqrtx)+e^(5x))dx#,

#=int(1/(x*x^(1/2))+e^(5x))dx#,

#=int(1/(x^(3/2))+e^(5x))dx#,

#=int(x^(-3/2)+e^(5x))dx#,

#=intx^(-3/2)dx+inte^(5x)dx#,

Here, we use, #intx^ndx=x^(n+1)/(n+1)+c, n!=-1#, while,

#inte^(5x)dx=1/5e^(5x)+c'#, as in Part (a) above.

#:. I_(b)=x^(-3/2+1)/(-3/2+1)+1/5e^(5x)#,

# rArr I_(b)=1/5e^(5x)-2/sqrtx+C_(b)#.

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