Problem 13
Question
Find the integral. (Note: Solve by the simplest method-not all require integration by parts.) $$ \int t \ln (t+1) d t $$
Step-by-Step Solution
Verified Answer
The integral \(\int t \ln(t+1) dt = \frac{1}{2} t^{2} \ln(t+1) - \frac{1}{4} t^{2} + \frac{1}{2} t - \frac{1}{2} \ln|t+1| + C\).
1Step 1: Identify and Assign \(u\) and \(dv\)
We are following the LIATE rule for selecting \(u\) and \(dv\). Let \(u = \ln(t+1)\) and \(dv = t dt\).
2Step 2: Calculate \(du\) and \(v\)
Differentiate \(u\) to find \(du\) and integrate \(dv\) to find \(v\). So, \(du = \frac{1}{t+1} dt\) and \(v = \frac{1}{2} t^{2}\).
3Step 3: 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: \(\int t \ln(t+1) dt = \frac{1}{2} t^{2} \ln(t+1) - \int \frac{1}{2} t^{2} \frac{1}{t+1} dt\).
4Step 4: 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.
Other exercises in this chapter
Problem 13
Use the Special Integration Formulas (Theorem 6.2) to find the integral. $$ \int \sqrt{25-4 x^{2}} d x $$
View solution Problem 13
Use partial fractions to find the integral. $$ \int \frac{x^{2}+3 x-4}{x^{3}-4 x^{2}+4 x} d x $$
View solution Problem 14
Determine whether the improper integral diverges or converges. Evaluate the integral if it converges. $$ \int_{1}^{\infty} \frac{4}{\sqrt[4]{x}} d x $$
View solution Problem 14
Find the integral. $$ \int x^{2} \sin ^{2} x d x $$
View solution