Evaluate the integral: I = integral dx/(sqrt(x² + 7))
Evaluate the integral: $$ I = \int \frac{dx}{\sqrt{x^2 + 7}}. $$
Explanation of Method
The form $\sqrt{x^2+a^2}$ under a radical calls for the tangent substitution $x=a\tan\theta$, since $1+\tan^2\theta=\sec^2\theta$ removes the square root. As in the secant-substitution case, the integral reduces to $\int \sec\theta\,d\theta$, which is then rewritten back in terms of $x$.
- Substitute $x=a\tan\theta$ to match the radical to a Pythagorean identity.
- Simplify the square root and $dx$ in terms of $\theta$.
- Integrate the resulting $\sec\theta$ using its standard antiderivative.
- Back-substitute to express the answer in terms of $x$, taking care with logarithms of a quotient.
Step 1 · Use a standard trigonometric substitution
We recognize that the integrand is of the form $\displaystyle \int \frac{dx}{\sqrt{x^2+a^2}}$, which suggests using the trigonometric substitution:
$$ x = \sqrt{7}\tan(\theta). $$
This substitution works because $\sec^2(\theta)=1+\tan^2(\theta)$, which will simplify the square root.
Step 2 · Find expressions for $dx$ and the square root
Using the substitution $x=\sqrt{7}\tan(\theta)$, we differentiate to find $dx$:
$$ dx = \sqrt{7}\sec^2(\theta)\,d\theta. $$
Next, we simplify the square root term:
$$ \sqrt{x^2+7} = \sqrt{7\tan^2(\theta)+7} = \sqrt{7(\tan^2(\theta)+1)} = \sqrt{7\sec^2(\theta)} = \sqrt{7}\sec(\theta). $$
Step 3 · Substitute into the integral
Substituting $x=\sqrt7\tan(\theta)$, $dx=\sqrt7\sec^2(\theta)\,d\theta$, and $\sqrt{x^2+7}=\sqrt7\sec(\theta)$ into the original integral:
$$ I = \int \frac{\sqrt{7}\sec^2(\theta)\,d\theta}{\sqrt{7}\sec(\theta)}. $$
Simplifying:
$$ I = \int \sec(\theta)\,d\theta. $$
Step 4 · Integrate
The integral of $\sec(\theta)$ is a standard result:
$$ \int \sec(\theta)\,d\theta = \ln\left|\sec(\theta)+\tan(\theta)\right| + C. $$
Step 5 · Back-substitute
We now need to back-substitute using $x=\sqrt7\tan(\theta)$. From this, we get:
$$ \tan(\theta) = \frac{x}{\sqrt{7}}. $$
Also, using the identity $\sec^2(\theta)=1+\tan^2(\theta)$, we find:
$$ \sec(\theta) = \sqrt{1+\left(\frac{x}{\sqrt{7}}\right)^2} = \sqrt{1+\frac{x^2}{7}} = \frac{\sqrt{x^2+7}}{\sqrt{7}}. $$
Thus, the integral becomes:
$$ I = \ln\left|\frac{\sqrt{x^2+7}}{\sqrt{7}}+\frac{x}{\sqrt{7}}\right| + C = \ln\left|\frac{x+\sqrt{x^2+7}}{\sqrt{7}}\right| + C. $$
Since $\ln\left|\dfrac{x+\sqrt{x^2+7}}{\sqrt7}\right| = \ln\left|x+\sqrt{x^2+7}\right| - \ln\sqrt7$, and the constant $-\ln\sqrt7$ can be absorbed into $C$, the antiderivative simplifies to:
$$ I = \ln\left|x+\sqrt{x^2+7}\right| + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Substitution | $x = \sqrt7\tan\theta$ |
| $\sqrt{x^2+7}$ | $\sqrt7\sec\theta$ |
| Reduced integral | $\int \sec\theta\,d\theta$ |
Additional Notes
- The integrand is defined for all real $x$, since $x^2+7 > 0$ always.
- A common slip here is rewriting $\ln\left|\dfrac{x+\sqrt{x^2+7}}{\sqrt7}\right|$ as $\dfrac{1}{\sqrt7}\ln\left|x+\sqrt{x^2+7}\right|$ — that step is invalid, since dividing inside a logarithm is not the same as dividing outside it. The correct simplification uses $\ln(A/B)=\ln A-\ln B$, and the resulting constant $-\ln\sqrt7$ simply merges into $C$.
- The result can be checked by differentiating $\ln\left|x+\sqrt{x^2+7}\right|$ and confirming it returns $\dfrac{1}{\sqrt{x^2+7}}$.
Result
The solution to the integral is: