/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 75 Find the integrals. $$\int \fr... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the integrals. $$\int \frac{t^{2}+t}{\sqrt{t+1}} d t$$

Short Answer

Expert verified
The integral is \( \frac{(t+1)^3}{3} - (t+1)^2 + \frac{2(t+1)^{3/2}}{3} + (t+1) - 2\sqrt{t+1} + C \).

Step by step solution

01

Simplify the Integrand

We begin by simplifying the integrand \( \frac{t^2 + t}{\sqrt{t+1}} \). We can separate the terms in the numerator: \( \frac{t^2}{\sqrt{t+1}} + \frac{t}{\sqrt{t+1}} \).
02

Use Substitution

Choose substitution \( u = \sqrt{t+1} \), which implies \( u^2 = t+1 \) and \( t = u^2 - 1 \). Then differentiate both sides: \( 2u du = dt \). The expression \( \frac{t^2}{\sqrt{t+1}} \) becomes \( \frac{(u^2 - 1)^2}{u} \) and \( \frac{t}{\sqrt{t+1}} \) becomes \( \frac{u^2 - 1}{u} \).
03

Simplify and Rewrite Integrals

First integral: \( \int \frac{(u^2 - 1)^2}{u} \cdot 2u du = \int 2u(u^2 - 1)^2 du \).Second integral: \( \int \frac{u^2 - 1}{u} \cdot 2u du = \int 2(u^2 - 1) du \).
04

Expand and Integrate

Expand the first integral: \( 2\int (u^4 - 2u^2 + 1)u du = 2\int (u^5 - 2u^3 + u) du \).Expand the second integral directly: \( 2\int (u^2 - 1) du \).Now integrate each term separately:1. \( 2\left( \frac{u^6}{6} - \frac{2u^4}{4} + \frac{u^2}{2} \right) \).2. \( 2\left( \frac{u^3}{3} - u \right) \).
05

Combine Results and Simplify

Combine the results of both integrals: \[ \left( \frac{u^6}{3} - u^4 + u^2 \right) + \left( \frac{2u^3}{3} - 2u \right) \].Simplify: \[ \frac{u^6}{3} - u^4 + \frac{2u^3}{3} + u^2 - 2u + C \].
06

Substitute Back

Substitute \( u = \sqrt{t+1} \) back into the equation:\[ \frac{(t+1)^3}{3} - (t+1)^2 + \frac{2(t+1)^{3/2}}{3} + (t+1) - 2\sqrt{t+1} + C \]This is the complete expression for the integrated function.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration by Substitution
Integration by substitution is a technique that simplifies complex integrals by transforming variables. Imagine it as changing clothes to suit a different weather. It's about switching one variable to another that makes the problem easier to solve.

For complex expressions, like our integrand \[\int \frac{t^2 + t}{\sqrt{t+1}} dt\]integration by substitution can simplify the process. Here's a neat trick: we use substitution to replace a complex part with a single letter, such as using \( u \) instead of \( \sqrt{t+1} \). This turns our challenging equation into something far more manageable.
  • First, decide what part of the integrand to substitute. Here, \( u = \sqrt{t+1} \).
  • Differentiate to find \( dt \) in terms of \( du \).
  • Rewrite the integral with \( u \).
This substitution simplifies the integrand and often converts it into a polynomial form, making it possible to integrate each term separately.
Polynomial Integration
Polynomial integration is the process of integrating expressions where the variables have been expanded into polynomial terms. It involves straightforward rules that turn polynomial expressions into antiderivatives easily.

Once the integral has been rewritten using the substitution as a polynomial, each term such as \( u^5, u^3, \) or \( u \) can be integrated individually. Here’s how it works:
  • For any term \( x^n \), it integrates to \( \frac{x^{n+1}}{n+1} \).
  • Apply this operation to each polynomial term separately.
So, when you have an equation like: \[2\int (u^5 - 2u^3 + u) du\]you simply integrate each part, applying the polynomial rule. This makes polynomial integration a breeze since you're often dealing with simple addition and multiplication operations.
Rational Expressions
Rational expressions involve ratios of polynomials and require a specific approach during integration. These expressions typically look complex at first, resembling the form \( \frac{P(x)}{Q(x)} \), where both the numerator \( P(x) \) and the denominator \( Q(x) \) are polynomials.

Turning our original function into a polynomial rational expression allows integration techniques to simplify each individual term. In both steps of the exercise, we split and simplified \[\frac{t^2 + t}{\sqrt{t+1}} \]using the substitution technique.
  • The substitution rewritten forms lead us to polynomial terms.
  • Simplifying the terms into polynomial format directly relates to integration prowess.
Once expanded and rewritten, these transformations enable easier integration through polynomial techniques.
Definite and Indefinite Integrals
Definite and indefinite integrals are fundamental in understanding integration. Let's clarify the difference:

An **indefinite integral** represents a family of functions and includes a constant \( C \), symbolizing all possible antiderivatives. When you see expressions like:
\[\int 2x dx = x^2 + C\]it's demonstrating an indefinite integral. In our exercise, the solution is indefinite since it ends with a \( + C \).

On the other hand, a **definite integral** calculates the specific area under a curve between two points \([a, b]\). Its outcome is a precise numerical value. If our problem involved setting specific limits (like from 1 to 3), the outcome would yield an exact answer without the constant \( C \).
  • Indefinite integrals tell us possible solutions; everything you need about the general form of the function.
  • Definite integrals give us exact areas, capturing true numerical value.
Knowing the purpose and application of each can deepen your comprehension of calculus and how integration suits various needs.

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