Trig substitution for a shifted quadratic radical

Key takeaway: This integral is designed for a substitution that simplifies the quadratic expression inside the square root. The numerator is closely related to the derivative of the radicand, which is the main clue.

Spot the structure

We want to evaluate

$\int \frac{2x+5}{\sqrt{x^2-2x+10}}\,dx.$

The expression inside the square root is a quadratic. Its derivative is $2x-2$, which is close to the numerator $2x+5$. That tells us to rewrite the numerator in a way that separates a derivative part from a constant part.

First complete the square:

$x^2-2x+10=(x-1)^2+9.$

This is useful because it shows the radicand is always positive and suggests a shift substitution.

Split the integrand into a useful form

Write

$2x+5=(2x-2)+7.$

Then the integral becomes

$\int \frac{2x-2}{\sqrt{x^2-2x+10}}\,dx + 7\int \frac{1}{\sqrt{x^2-2x+10}}\,dx.$

The first part is immediate with the substitution

$u=x^2-2x+10,\qquad du=(2x-2)dx.$

So

$\int \frac{2x-2}{\sqrt{x^2-2x+10}}\,dx = \int u^{-1/2}\,du = 2\sqrt{u}=2\sqrt{x^2-2x+10}.$

Handle the remaining standard form

For the second part, use $x-1=3\tan\theta$ or recognize the standard formula

$\int \frac{dx}{\sqrt{(x-1)^2+3^2}} = \ln\left|x-1+\sqrt{(x-1)^2+9}\right|+C.$

So

$7\int \frac{1}{\sqrt{x^2-2x+10}}\,dx = 7\ln\left|x-1+\sqrt{x^2-2x+10}\right|+C.$

Final answer

Putting both parts together,

$\boxed{\int \frac{2x+5}{\sqrt{x^2-2x+10}}\,dx = 2\sqrt{x^2-2x+10}+7\ln\left|x-1+\sqrt{x^2-2x+10}\right|+C}.$

Why this works

The numerator was intentionally close to the derivative of the quadratic. That makes decomposition the fastest path. Completing the square then reveals a shifted square-root form, which is one of the most common antiderivatives in calculus.

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Pitfalls the pros know 👇 A frequent mistake is trying to force a single substitution on the whole numerator without checking whether the numerator matches the derivative of the radicand. Here it does not match exactly, so you must split $2x+5$ into a derivative-like piece plus a constant remainder. Another error is forgetting the absolute value inside the logarithm. For square-root integrals of the form $\sqrt{(x-a)^2+b^2}$, the logarithmic antiderivative needs the absolute value.

What if the problem changes? If the integral were $\int \frac{2x-2}{\sqrt{x^2-2x+10}}\,dx$, then it would collapse to a direct substitution with $u=x^2-2x+10$ and the answer would simply be $2\sqrt{x^2-2x+10}+C$. If the numerator were a pure constant, the antiderivative would instead be the standard logarithmic form. The difficulty changes depending on how closely the numerator matches the derivative of the quadratic.

Tags: complete the square, radical substitution, standard logarithmic integral