Questions: Evaluate the following integral. [ int fracxsqrtx+9 d x ] E to view the table of general integration formulas. [ int fracxsqrtx+9 d x=square ]

$Evaluate the following integral. [ int fracxsqrtx+9 d x ] E to view the table of general integration formulas. [ int fracxsqrtx+9 d x=square ]$

Transcript text: Evaluate the following integral. \[ \int \frac{x}{\sqrt{x+9}} d x \] E to view the table of general integration formulas. \[ \int \frac{x}{\sqrt{x+9}} d x=\square \]

Solution

Solution Steps

To evaluate the integral $\int \frac{x}{\sqrt{x+9}} \, dx$, we can use a substitution method. Let $u = x + 9$, then $du = dx$ and $x = u - 9$. Substitute these into the integral and simplify.

Step 1: Substitution

To evaluate the integral $\int \frac{x}{\sqrt{x+9}} \, dx$, we use the substitution $u = x + 9$. Then, $du = dx$ and $x = u - 9$.

Step 2: Substitute and Simplify

Substitute $u$ and $du$ into the integral: \[ \int \frac{x}{\sqrt{x+9}} \, dx = \int \frac{u-9}{\sqrt{u}} \, du \] This simplifies to: \[ \int \frac{u-9}{\sqrt{u}} \, du = \int \left( \frac{u}{\sqrt{u}} - \frac{9}{\sqrt{u}} \right) du = \int \left( \sqrt{u} - \frac{9}{\sqrt{u}} \right) du \]

Step 3: Integrate

Now, integrate each term separately: \[ \int \sqrt{u} \, du - 9 \int \frac{1}{\sqrt{u}} \, du \] Using the power rule for integration: \[ \int u^{1/2} \, du = \frac{2}{3} u^{3/2} \] \[ \int u^{-1/2} \, du = 2 u^{1/2} \] Thus, the integral becomes: \[ \frac{2}{3} u^{3/2} - 9 \cdot 2 u^{1/2} = \frac{2}{3} u^{3/2} - 18 u^{1/2} \]

Step 4: Substitute Back

Substitute $u = x + 9$ back into the expression: \[ \frac{2}{3} (x+9)^{3/2} - 18 (x+9)^{1/2} \]