Evaluate the integral: I = integral x/(sqrt(2x + 5)) dx
Evaluate the integral: $$ I = \int \frac{x}{\sqrt{2x + 5}} \, dx. $$
Explanation of Method
When $x$ itself (not just a function of the linear expression) sits in the numerator, the substitution $u=2x+5$ requires also solving for $x$ in terms of $u$ so the numerator can be rewritten entirely in $u$. The integrand then splits into two power-rule terms.
- Substitute $u$ for the linear expression under the root.
- Solve for $x$ in terms of $u$ to rewrite the numerator.
- Split the resulting fraction into two power terms and integrate each.
- Substitute back and add the constant of integration.
Step 1 · Use substitution
To simplify this integral, we use the substitution method. Let:
$$ u = 2x + 5. $$
Then, the derivative of $u$ with respect to $x$ is:
$$ \frac{du}{dx} = 2 \quad \Rightarrow \quad du = 2\,dx \quad \Rightarrow \quad dx = \frac{du}{2}. $$
Also, solve for $x$ in terms of $u$:
$$ u = 2x+5 \quad \Rightarrow \quad x = \frac{u-5}{2}. $$
Now substitute into the integral:
$$ I = \int \frac{\frac{u-5}{2}}{\sqrt{u}} \cdot \frac{du}{2}. $$
Step 2 · Simplify the integral
The integral becomes:
$$ I = \int \frac{(u-5)}{4\sqrt{u}}\,du. $$
Now, split the integral into two parts:
$$ I = \frac{1}{4}\int \frac{u}{\sqrt{u}}\,du - \frac{5}{4}\int \frac{1}{\sqrt{u}}\,du. $$
Step 3 · Integrate each term
First term:
$$ \int \frac{u}{\sqrt{u}}\,du = \int u^{1/2}\,du = \frac{2}{3}u^{3/2}. $$
Second term:
$$ \int \frac{1}{\sqrt{u}}\,du = \int u^{-1/2}\,du = 2u^{1/2}. $$
Step 4 · Substitute back $u = 2x + 5$
Substitute $u = 2x+5$ back into the results:
$$ I = \frac{1}{4}\left(\frac{2}{3}(2x+5)^{3/2}\right) - \frac{5}{4}\left(2(2x+5)^{1/2}\right). $$
Simplify:
$$ I = \frac{1}{6}(2x+5)^{3/2} - \frac{5}{2}(2x+5)^{1/2} + C. $$
Summary Table
| Term | Antiderivative |
|---|---|
| $\tfrac{1}{4}u/\sqrt{u}$ | $\tfrac{1}{6}(2x+5)^{3/2}$ |
| $-\tfrac{5}{4}\cdot 1/\sqrt{u}$ | $-\tfrac{5}{2}(2x+5)^{1/2}$ |
Additional Notes
- The domain of $I$ requires $x > -\tfrac{5}{2}$, since $2x+5>0$ is needed for the square root in the denominator to be real and nonzero.
- Solving for $x=\tfrac{u-5}{2}$ so the numerator can be expressed purely in $u$ is the key extra step that distinguishes this problem from the simpler linear-substitution problems seen earlier — whenever $x$ appears outside the substituted expression, this step is required.
- The result can be checked by differentiating $\tfrac16(2x+5)^{3/2}-\tfrac52(2x+5)^{1/2}$ with the chain rule and confirming it returns $\dfrac{x}{\sqrt{2x+5}}$.
Result
The solution to the integral is: