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How do you integrate by substitution #int x^3(x^4+3)^2 dx#?

Answer 1

Christopher Allers

Please see the explanation.

Let #u = x^4 + 3#, then #du = 4x^3dx# or #x^3dx = (1/4)du#

#intx^3(x^4+3)^2dx = 1/4intu^2du#

Integrate using the power rule:

#intx^3(x^4+3)^2dx = (1/12)u^3 + C#

Reverse the substitution:

#intx^3(x^4+3)^2dx = (1/12)(x^4 + 3)^3 + C#

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Answer 2

Charles Arocha

To integrate ( \int x^3(x^4+3)^2 ) with respect to ( x ), you can use the substitution method. Let ( u = x^4 + 3 ). Then, ( du = 4x^3 dx ).

We solve for ( dx ): ( dx = \frac{du}{4x^3} ).

Now, rewrite the integral in terms of ( u ):

[ \int x^3(x^4+3)^2 dx = \int \frac{1}{4} u^2 du ]

Integrating ( \frac{1}{4} u^2 ) with respect to ( u ), we get:

[ \frac{1}{4} \cdot \frac{u^3}{3} + C ]

Substitute back for ( u ):

[ \frac{1}{4} \cdot \frac{(x^4 + 3)^3}{3} + C ]

So, the integral of ( x^3(x^4+3)^2 ) with respect to ( x ) is ( \frac{1}{4} \cdot \frac{(x^4 + 3)^3}{3} + C ), where ( C ) is the constant of integration.

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