Problem

Integrate Using u-Substitution integral of x square root of 1+x with respect to x

The problem asks for the evaluation of a definite or indefinite integral using the technique of u-substitution. In calculus, u-substitution, also known as variable substitution, is used to simplify integrals by changing the variable being integrated against to a new variable that simplifies the integral.

Specifically, for this problem, the integral to solve is:

∫ x√(1 + x) dx

You are supposed to perform u-substitution by identifying an appropriate function of x (let's call it u) that, when substituted, makes the integral easier to solve. This typically involves finding a portion of the integrand whose derivative is also present elsewhere in the integrand.

$\int x \sqrt{1 + x} d x$

Answer

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Solution:

Step 1

Choose $u = 1 + x$ and calculate $du$.

Step 1.1

Set $u = 1 + x$ and determine $\frac{du}{dx}$.

Step 1.1.1

Differentiate $1 + x$ to find $\frac{d}{dx} (1 + x)$.

Step 1.1.2

Utilize the Sum Rule: the derivative of $1 + x$ is the sum of the derivatives $\frac{d}{dx} (1) + \frac{d}{dx} (x)$.

Step 1.1.3

The derivative of a constant is zero, so $\frac{d}{dx} (1) = 0$ and we have $0 + \frac{d}{dx} (x)$.

Step 1.1.4

Apply the Power Rule: $\frac{d}{dx} (x^n) = n x^{n-1}$ for $n = 1$, which gives us $0 + 1$.

Step 1.1.5

Combine the terms to get $du = dx$.

Step 1.2

Substitute $u$ and $du$ into the integral: $\int (u - 1) \sqrt{u} du$.

Step 2

Express $\sqrt{u}$ as $u^{\frac{1}{2}}$: $\int (u - 1) u^{\frac{1}{2}} du$.

Step 3

Expand the integrand $(u - 1) u^{\frac{1}{2}}$.

Step 3.1

Distribute $u^{\frac{1}{2}}$ across $u - 1$: $\int u u^{\frac{1}{2}} - 1 \cdot u^{\frac{1}{2}} du$.

Step 3.2

Raise $u$ to the first power: $\int u^1 u^{\frac{1}{2}} - 1 \cdot u^{\frac{1}{2}} du$.

Step 3.3

Combine the exponents using $a^m a^n = a^{m+n}$: $\int u^{1 + \frac{1}{2}} - 1 \cdot u^{\frac{1}{2}} du$.

Step 3.4

Express $1$ as a fraction: $\int u^{\frac{2}{2} + \frac{1}{2}} - 1 \cdot u^{\frac{1}{2}} du$.

Step 3.5

Add the exponents over a common denominator: $\int u^{\frac{3}{2}} - 1 \cdot u^{\frac{1}{2}} du$.

Step 3.6

Simplify the expression: $\int u^{\frac{3}{2}} - u^{\frac{1}{2}} du$.

Step 4

Separate the integral into two parts: $\int u^{\frac{3}{2}} du - \int u^{\frac{1}{2}} du$.

Step 5

Integrate $u^{\frac{3}{2}}$ using the Power Rule: $\frac{2}{5} u^{\frac{5}{2}} + C_1 - \int u^{\frac{1}{2}} du$.

Step 6

Factor out the constant $-1$: $\frac{2}{5} u^{\frac{5}{2}} + C_1 - \int u^{\frac{1}{2}} du$.

Step 7

Integrate $u^{\frac{1}{2}}$ using the Power Rule: $\frac{2}{5} u^{\frac{5}{2}} + C_1 - \left( \frac{2}{3} u^{\frac{3}{2}} + C_2 \right)$.

Step 8

Combine constants and simplify: $\frac{2}{5} u^{\frac{5}{2}} - \frac{2}{3} u^{\frac{3}{2}} + C$.

Step 9

Substitute back $1 + x$ for $u$: $\frac{2}{5} (1 + x)^{\frac{5}{2}} - \frac{2}{3} (1 + x)^{\frac{3}{2}} + C$.

Knowledge Notes:

To solve the integral of $x \sqrt{1+x}$ using $u$-substitution, we follow these steps:

  1. $u$-Substitution: This technique involves choosing a substitution $u = g(x)$ that simplifies the integral. The differential $du$ is then the derivative of $u$ with respect to $x$, multiplied by $dx$.

  2. Sum Rule in Differentiation: The derivative of a sum of functions is the sum of the derivatives of those functions.

  3. Constant Rule in Differentiation: The derivative of a constant is zero.

  4. Power Rule in Differentiation: For any real number $n$, the derivative of $x^n$ with respect to $x$ is $n x^{n-1}$.

  5. Distributive Property: This property allows us to multiply a sum by multiplying each addend separately and then sum the products.

  6. Combining Exponents: When multiplying like bases, we add the exponents: $a^m \cdot a^n = a^{m+n}$.

  7. Power Rule in Integration: The integral of $u^n$ with respect to $u$ is $\frac{u^{n+1}}{n+1} + C$, provided $n \neq -1$.

  8. Simplifying Expressions: Combining like terms and simplifying expressions are important to reach the final form of the integral.

  9. Back-Substitution: After integrating with respect to $u$, we substitute back the original variable to express the antiderivative in terms of the original variable.

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