Find the indefinite integral.
$\int \frac{x}{\sqrt{x + 3}} d x$
$\int \frac{x}{\sqrt{x + 3}} d x = ◻$

Answer

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Answer

$\frac{2}{3} (x + 3)^{3 / 2} - 6 (x + 3)^{1 / 2} + C$ is the indefinite integral of $\int \frac{x}{\sqrt{x + 3}} d x$

Steps

Step 1 ：Let $u = x + 3$ . Then, $d u = d x$ and $x = u - 3$ .

Step 2 ：Substitute these values into the integral: $\int \frac{x}{\sqrt{x + 3}} d x = \int \frac{u - 3}{\sqrt{u}} d u$

Step 3 ：Split the fraction into two parts: $\int \frac{u - 3}{\sqrt{u}} d u = \int \frac{u}{\sqrt{u}} d u - \int \frac{3}{\sqrt{u}} d u$

Step 4 ：Simplify the integrals: $\int \frac{u}{\sqrt{u}} d u - \int \frac{3}{\sqrt{u}} d u = \int u^{1 / 2} d u - 3 \int u^{- 1 / 2} d u$

Step 5 ：Now, we can integrate: $\int u^{1 / 2} d u - 3 \int u^{- 1 / 2} d u = \frac{2}{3} u^{3 / 2} - 6 u^{1 / 2} + C$

Step 6 ：Substitute $u = x + 3$ back into the equation: $\frac{2}{3} u^{3 / 2} - 6 u^{1 / 2} + C = \frac{2}{3} (x + 3)^{3 / 2} - 6 (x + 3)^{1 / 2} + C$

Step 7 ： $\frac{2}{3} (x + 3)^{3 / 2} - 6 (x + 3)^{1 / 2} + C$ is the indefinite integral of $\int \frac{x}{\sqrt{x + 3}} d x$

Find the indefinite integral.∫xx+3dx∫xx+3dx=◻

Answer

Popular Problems

Find the indefinite integral.
$\int \frac{x}{\sqrt{x + 3}} d x$
$\int \frac{x}{\sqrt{x + 3}} d x = ◻$