Consider the following integral:

[tex]\[ \int \frac{x^3}{x^4-4} \, dx \][/tex]

Given the substitution [tex]\( u = x^4 - 4 \)[/tex], find [tex]\( du \)[/tex].

[tex]\[ du = (\square) \, dx \][/tex]

Rewrite the given integral in terms of [tex]\( u \)[/tex].

[tex]\[ \int (\square) \, du \][/tex]

Evaluate the integral by making the given substitution. (Remember to use absolute values where appropriate. Use [tex]\( C \)[/tex] for the constant of integration.)

[tex]\[ \square \][/tex]



Answer :

To solve the integral:

[tex]\[ \int \frac{x^3}{x^4-4} \, dx \][/tex]

we use the given substitution [tex]\( u = x^4 - 4 \)[/tex].

1. Find [tex]\( du \)[/tex] in terms of [tex]\( x \)[/tex] and [tex]\( dx \)[/tex]:

We start by differentiating [tex]\( u \)[/tex] with respect to [tex]\( x \)[/tex]:

[tex]\[ u = x^4 - 4 \][/tex]
[tex]\[ \frac{du}{dx} = 4x^3 \][/tex]

Therefore,

[tex]\[ du = 4x^3 \, dx \][/tex]

Rearranging to solve for [tex]\( dx \)[/tex],

[tex]\[ dx = \frac{du}{4x^3} \][/tex]

2. Rewrite the integral in terms of [tex]\( u \)[/tex]:

Substitute [tex]\( u = x^4 - 4 \)[/tex] and [tex]\( dx = \frac{du}{4x^3} \)[/tex] into the original integral:

[tex]\[ \int \frac{x^3}{x^4 - 4} \, dx = \int \frac{x^3}{u} \cdot \frac{du}{4x^3} \][/tex]

Simplify the integral:

[tex]\[ \int \frac{1}{u} \cdot \frac{du}{4} = \frac{1}{4} \int \frac{1}{u} \, du \][/tex]

3. Evaluate the integral:

The integral [tex]\( \int \frac{1}{u} \, du \)[/tex] is a standard integral:

[tex]\[ \int \frac{1}{u} \, du = \ln|u| + C \][/tex]

Therefore,

[tex]\[ \frac{1}{4} \int \frac{1}{u} \, du = \frac{1}{4} \ln|u| + C \][/tex]

Substitute back [tex]\( u = x^4 - 4 \)[/tex]:

[tex]\[ \frac{1}{4} \ln|x^4 - 4| + C \][/tex]

Thus, the evaluated integral is:

[tex]\[ \int \frac{x^3}{x^4 - 4} \, dx = \frac{1}{4} \ln|x^4 - 4| + C \][/tex]