Evaluate the integral: I = integral dx/(sqrt(2 - 5x))
Evaluate the integral: $$ I = \int \frac{dx}{\sqrt{2 - 5x}}. $$
Explanation of Method
The reciprocal square root $\frac{1}{\sqrt{2-5x}}$ is a negative fractional power, $(2-5x)^{-1/2}$, of a linear expression. Substituting $u=2-5x$ and applying the power rule handles it in the same routine way as the earlier linear-substitution problems.
- Substitute $u$ for the linear expression inside the root.
- Rewrite the reciprocal root as a negative fractional power of $u$.
- Integrate using the power rule.
- Substitute back and add the constant of integration.
Step 1 · Use substitution
To simplify the integral, we will use substitution. Let:
$$ u = 2 - 5x. $$
Then, differentiate $u$ with respect to $x$:
$$ \frac{du}{dx} = -5 \quad \Rightarrow \quad dx = \frac{du}{-5}. $$
Substituting into the integral, we get:
$$ I = \int \frac{1}{\sqrt{u}}\cdot \frac{du}{-5} = -\frac{1}{5}\int u^{-1/2}\,du. $$
Step 2 · Integrate
We now integrate $u^{-1/2}$. Using the power rule for integration:
$$ \int u^n\,du = \frac{u^{n+1}}{n+1} + C \quad \text{for } n\neq -1, $$
we get:
$$ \int u^{-1/2}\,du = \frac{u^{1/2}}{1/2} = 2u^{1/2}. $$
Step 3 · Substitute back $u = 2 - 5x$
Now, substitute back $u = 2-5x$ into the result:
$$ I = -\frac{1}{5}\cdot 2(2-5x)^{1/2}. $$
Step 4 · Add the constant of integration
Finally, we add the constant of integration $C$ to the result:
$$ I = -\frac{2}{5}(2-5x)^{1/2} + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Substitution | $u = 2-5x$ |
| $dx$ in terms of $du$ | $-\tfrac{1}{5}du$ |
| Antiderivative in $u$ | $2u^{1/2}$ |
Additional Notes
- The domain of $I$ requires $x < \tfrac{2}{5}$, since $2-5x > 0$ is needed for the square root in the denominator to be real and nonzero.
- The final coefficient $-\tfrac25$ comes from multiplying the substitution factor $-\tfrac15$ by the power-rule factor $2$.
- The result can be checked by differentiating $-\tfrac25(2-5x)^{1/2}$ with the chain rule and confirming it returns $(2-5x)^{-1/2}$.
Result
The solution to the integral is: