To solve the integral [tex]\(\int \frac{t}{t^4+81} \, dt\)[/tex], we can use a method involving a suitable substitution.
First, observe that the integrand [tex]\(\frac{t}{t^4 + 81}\)[/tex] suggests a substitution that simplifies the polynomial in the denominator. Let's perform the substitution:
[tex]\[ u = t^2 \][/tex]
Then,
[tex]\[ du = 2t \, dt \][/tex]
or equivalently,
[tex]\[ dt = \frac{du}{2t} \][/tex]
Since [tex]\( t^2 = u \)[/tex], we have [tex]\( t = \sqrt{u} \)[/tex]. Substituting these into our integral gives us:
[tex]\[ \int \frac{t}{t^4 + 81} \, dt = \int \frac{\sqrt{u}}{(\sqrt{u})^4 + 81} \cdot \frac{du}{2\sqrt{u}} \][/tex]
Simplifying inside the integrand, we get:
[tex]\[ \int \frac{\sqrt{u}}{u^2 + 81} \cdot \frac{du}{2\sqrt{u}} = \frac{1}{2} \int \frac{1}{u^2 + 81} \, du \][/tex]
Now, this integral can be recognized as the standard form [tex]\(\int \frac{1}{a^2 + u^2} \, du = \frac{1}{a} \arctan\left(\frac{u}{a}\right) + C\)[/tex], where [tex]\(a = 9\)[/tex] in our case since [tex]\(81 = 9^2\)[/tex].
Therefore, the integral becomes:
[tex]\[ \frac{1}{2} \int \frac{1}{u^2 + 9^2} \, du = \frac{1}{2} \cdot \frac{1}{9} \arctan\left(\frac{u}{9}\right) + C \][/tex]
Simplifying the constants outside the integral:
[tex]\[ \frac{1}{2} \cdot \frac{1}{9} \arctan\left(\frac{u}{9}\right) + C = \frac{1}{18} \arctan\left(\frac{u}{9}\right) + C \][/tex]
Finally, we revert back to the original variable [tex]\( t \)[/tex] by using [tex]\( u = t^2 \)[/tex], so our integral in terms of [tex]\( t \)[/tex] is:
[tex]\[ \frac{1}{18} \arctan\left(\frac{t^2}{9}\right) + C \][/tex]
Thus, the indefinite integral is:
[tex]\[ \int \frac{t}{t^4 + 81} \, dt = \frac{1}{18} \arctan\left(\frac{t^2}{9}\right) + C \][/tex]
