AP Calculus AB and BC

Integral of (2x+3)/(x^2+3x): Spot the Derivative

Q: What is the integral of (2x+3)/(x^2+3x)?

One substitution, u = x^{2}+3x, carries it all the way to \ln\left|x^{2}+3x\right| + C, with no coefficient out front.

Q: Is ln|x| + ln|x+3| the same answer?

Yes. The product law combines it into \ln\left|x^{2}+3x\right|, so the two forms are identical.

The integral of 2x plus 3 over x squared plus 3x is the natural log of the absolute value of x squared plus 3x, plus C. The numerator is exactly the derivative of the denominator, so the substitution u equals x squared plus 3x finishes the problem in one line with no coefficient in front.

\int \frac{2x+3}{x^{2}+3x}\,dx = \ln\left|x^{2}+3x\right| + C

Differentiate the denominator first

Before doing anything with a rational integrand, differentiate the bottom. Here $\frac{d}{dx}\left(x^{2}+3x\right) = 2x+3$ , which is the numerator, character for character. Any integrand shaped like $\frac{g'(x)}{g(x)}$ integrates to $\ln\left|g(x)\right|$ .

So take $u = x^{2}+3x$ and $du = (2x+3)\,dx$ . The substitution consumes the entire numerator, which is why no fraction appears out front.

\int \frac{2x+3}{x^{2}+3x}\,dx = \int \frac{du}{u} = \ln\left|u\right| + C = \ln\left|x^{2}+3x\right| + C

Partial fractions gets there too, slowly

The denominator does factor, as $x(x+3)$ , so a BC student can legitimately decompose it. Solving $2x+3 = A(x+3) + Bx$ gives $A = 1$ and $B = 1$ , and both logs come out with coefficient one.

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

Same answer, three times the writing

Decomposing is not an error here, it is wasted minutes and three extra chances to drop a sign. Check the numerator against the derivative of the denominator before you set up A and B. When they match, the answer is one logarithm and you are done.

The mistakes students make

The pattern is easy once seen, and the errors below all come from not looking for it.

Dividing by the derivative and writing $\frac{\ln\left|x^{2}+3x\right|}{2x+3}$ . Antiderivatives never divide by a non-constant function that way; only a constant can be pulled out.
Attaching a stray $\frac{1}{2}$ and answering $\frac{1}{2}\ln\left|x^{2}+3x\right|$ . That would be right if the numerator were $x + \frac{3}{2}$ , half of $du$ . As written, $du$ matches the numerator exactly and no coefficient is needed.
Breaking the fraction into $\int\frac{2x}{x^{2}+3x}dx + \int\frac{3}{x^{2}+3x}dx$ . Both halves do come out to logs, the first because $\frac{2x}{x^{2}+3x}$ cancels to $\frac{2}{x+3}$ , but you have turned one substitution into two problems plus a recombination step.

Every answer on this page is machine checked

An automated test differentiates the antiderivative above and confirms it returns the integrand. A wrong sign or a missing factor fails the build, so it cannot reach you.

Frequently asked questions

What is the integral of (2x+3)/(x^2+3x)?

One substitution, $u = x^{2}+3x$ , carries it all the way to $\ln\left|x^{2}+3x\right| + C$ , with no coefficient out front.

How do I know when a fraction integrates to a natural log?

When the numerator is the derivative of the denominator, or a constant multiple of it. Differentiate the bottom and compare; if the derivative of the bottom is $k$ times the numerator, the answer is $\frac{1}{k}\ln\left|g(x)\right|$ .

Is ln|x| + ln|x+3| the same answer?

Yes. The product law combines it into $\ln\left|x^{2}+3x\right|$ , so the two forms are identical.