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Chapter 4: Problem 8

Find the integral. $$ \int \frac{t}{t^{4}+16} d t $$

Short Answer

Expert verified
The integral equals to \(\frac{1}{4} \ln |t^{4} + 16| + C\).

Step by step solution

01

Identify the suitable substitution

Take a look at the integral, the derivative of \(t^{4}\) is \(4t^3\), which is somewhat similar to the numerator (up to a constant multiple). This implies that if we let \(u = t^{4}\), the numerator \(t dt\) will become \(\frac{1}{4}du\). So we rewrite the integral in terms of \(u\).
02

Perform the substitution and solve the integral

With substitution \(u = t^4\), we have \(du = 4t^3dt\). Therefore \( t dt = \frac{1}{4} du\). We substitute both into the original integral: \(\int \frac{t}{t^{4}+16} dt\) becomes \(\int \frac{1}{4} \cdot \frac{1}{u+16} du\). This now becomes a standard integral. By recognizing this as the integral of 1 over (u+16), we can integrate simply by applying the natural logarithm rule to find the antiderivative.
03

Substitute back and simplify

After integrating we have \(\frac{1}{4} \ln |u + 16|\). Substituting \(u = t^{4}\) back into the equation, we obtain \(\frac{1}{4} \ln |t^{4} + 16|\). Adding \(C\) for the constant of integration, we get \(\frac{1}{4} \ln |t^{4} + 16| + C\).

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