AP Calculus AB and BC

Integral of 1/(x+1)^2: Answer and Steps

The integral of 1/(x + 1)^2 is -1/(x + 1) + C. Substituting u = x + 1 turns the integrand into u^(-2), and the reverse power rule gives -u^(-1). No logarithm appears, because the exponent is -2 rather than -1, which is the only power the rule excludes.

∫1(x+1)2 dx=−1x+1+C\int \frac{1}{(x+1)^2}\,dx = -\frac{1}{x+1} + C

Substitute, then use the power rule

The squared denominator is the only awkward part, and it disappears under u=x+1u = x+1. That substitution is unusually clean: du=dxdu = dx with no leftover constant to balance.

∫1(x+1)2 dx=∫u−2 du\int \frac{1}{(x+1)^2}\,dx = \int u^{-2}\,du

The exponent −2-2 is not the excluded case, so the reverse power rule applies directly: add 1 to get −1-1, then divide by −1-1.

∫u−2 du=u−1−1+C=−1u+C\int u^{-2}\,du = \frac{u^{-1}}{-1} + C = -\frac{1}{u} + C
∫1(x+1)2 dx=−1x+1+C\int \frac{1}{(x+1)^2}\,dx = -\frac{1}{x+1} + C

Checking, and where the answer is valid

Rewrite the result as −(x+1)−1-(x+1)^{-1} and differentiate. The chain rule contributes an inside derivative of 1, so nothing else changes.

ddx[−(x+1)−1]=(x+1)−2=1(x+1)2\frac{d}{dx}\left[-(x+1)^{-1}\right] = (x+1)^{-2} = \frac{1}{(x+1)^2}

The integrand is undefined at x=−1x = -1, so an antiderivative only describes the area on an interval that stays on one side of that point.

A negative answer is the tell

Plugging endpoints across the break gives nonsense: from x=−2x = -2 to x=0x = 0 the formula returns −2-2, yet the integrand is positive everywhere it is defined. A positive integrand cannot produce a negative definite integral, so that result flags a divergent integral rather than an arithmetic slip.

The mistake students make

The dominant error is answering ln⁡∣(x+1)2∣\ln|(x+1)^2| or ln⁡∣x+1∣\ln|x+1|, transplanting the ∫1u du\int \frac{1}{u}\,du rule onto ∫1u2 du\int \frac{1}{u^2}\,du. A fraction bar does not summon a logarithm; only the exponent −1-1 does. Differentiating ln⁡∣x+1∣\ln|x+1| returns 1x+1\frac{1}{x+1}, not 1(x+1)2\frac{1}{(x+1)^2}.

The second is the sign. Dividing by the new exponent −1-1 is what makes the answer negative, and an answer of 1x+1\frac{1}{x+1} differentiates to −1(x+1)2-\frac{1}{(x+1)^2}, the integrand with the wrong sign.

Read the exponent before choosing the rule

∫1u du=ln⁡∣u∣+C\int \frac{1}{u}\,du = \ln|u| + C, while ∫1u2 du=−1u+C\int \frac{1}{u^2}\,du = -\frac{1}{u} + C. Only n=−1n = -1 produces a logarithm. Every other power stays a power.

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

Why is there no absolute value in the answer?

Absolute value appears with logarithms because ln⁡\ln needs a positive input. −1x+1-\frac{1}{x+1} is already defined for every x≠−1x \neq -1, so there is no domain to repair.

Do I have to substitute at all?

No. The inside x+1x+1 has derivative 1, so you can apply the power rule straight to (x+1)−2(x+1)^{-2}. Substitution becomes worth writing out when the inside carries a coefficient, as in (2x+1)−2(2x+1)^{-2}, where a factor of 12\frac{1}{2} has to appear.

What is the integral of 1/(x+1)?

That is the excluded case n=−1n = -1, and it gives ln⁡∣x+1∣+C\ln|x+1| + C. The exponent decides which rule applies, not the fact that the expression is a fraction.