Question

How do you integrate A and B?

Answer 1

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`

Answer 2

`(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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