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Chapter 6: Problem 13

Find the integral. (Note: Solve by the simplest method-not all require integration by parts.) $$ \int t \ln (t+1) d t $$

Short Answer

Expert verified
The integral \(\int t \ln(t+1) dt = \frac{1}{2} t^{2} \ln(t+1) - \frac{1}{4} t^{2} + \frac{1}{2} t - \frac{1}{2} \ln|t+1| + C\).

Step by step solution

01

Identify and Assign \(u\) and \(dv\)

We are following the LIATE rule for selecting \(u\) and \(dv\). Let \(u = \ln(t+1)\) and \(dv = t dt\).
02

Calculate \(du\) and \(v\)

Differentiate \(u\) to find \(du\) and integrate \(dv\) to find \(v\). So, \(du = \frac{1}{t+1} dt\) and \(v = \frac{1}{2} t^{2}\).
03

Apply Integration by Parts Formula

Now, apply the formula for integration by parts, which is \(\int u dv = uv - \int v du\). After replacing \(u,v,du\), and \(dv\) with their values, the equation becomes: \(\int t \ln(t+1) dt = \frac{1}{2} t^{2} \ln(t+1) - \int \frac{1}{2} t^{2} \frac{1}{t+1} dt\).
04

Simplify the Integral and Solve

Simplify the integral to get \(\int t \ln(t+1) dt = \frac{1}{2} t^{2} \ln(t+1) - \int \frac{t}{2(t+1)} dt\). Now integrate the simplified part \(\int \frac{t}{2(t+1)} dt\) . After evaluating, the answer is \(\frac{1}{4} t^{2} - \frac{1}{2} t + \frac{1}{2} \ln|t+1| + C\), where C is the constant of integration.

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