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Integral of xx2+1\frac{x}{x^{2} + 1}

The integral of xx2+1\frac{x}{x^{2} + 1} is ln⁡(x2+1)2+C\frac{\ln\left(x^{2} + 1\right)}{2} + C.

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Graph of f(x) = x / (x^2 + 1) with the area from 0 to 1 shaded-0.4-0.20.20.40.60.811.21.41.6-0.4-0.20.20.40.6f(x)
Graph of f(x) = x / (x^2 + 1) with the area from 0 to 1 shaded. The shaded area equals the definite integral, about 0.34657359.

How to work it out

  1. 1

    We want an antiderivative of f(x) = x / (x2x^{2} + 1).

    ∫xx2+1 dx\int \frac{x}{x^{2} + 1}\,dx
  2. 2

    Method: substitution u = x² + 1.

  3. 3

    That gives the antiderivative; add the constant C because any constant disappears when you differentiate:

    ∫xx2+1 dx=ln⁡(x2+1)2+C\int \frac{x}{x^{2} + 1}\,dx = \frac{\ln\left(x^{2} + 1\right)}{2} + C
  4. 4

    Check by differentiating the answer; you must get the original function back:

    ddx[ln⁡(x2+1)2]=22xx2+14\frac{d}{dx}\left[\frac{\ln\left(x^{2} + 1\right)}{2}\right] = \frac{2 \frac{2 x}{x^{2} + 1}}{4}
  5. 5

    As a definite integral, evaluate the antiderivative at both ends and subtract. Between 0 and 1:

    ∫01xx2+1 dx=[ln⁡(x2+1)2]01≈0.34657359\int_{0}^{1} \frac{x}{x^{2} + 1}\,dx = \Big[\frac{\ln\left(x^{2} + 1\right)}{2}\Big]_{0}^{1} \approx 0.34657359
LaTeX source of the formulas on this page
\[ \int \frac{x}{x^{2} + 1}\,dx = \frac{\ln\left(x^{2} + 1\right)}{2} + C \]
\[ \int \frac{x}{x^{2} + 1}\,dx \]
\[ \frac{d}{dx}\left[\frac{\ln\left(x^{2} + 1\right)}{2}\right] = \frac{2 \frac{2 x}{x^{2} + 1}}{4} \]
\[ \int_{0}^{1} \frac{x}{x^{2} + 1}\,dx = \Big[\frac{\ln\left(x^{2} + 1\right)}{2}\Big]_{0}^{1} \approx 0.34657359 \]

Definite integrals of the same function

IntervalValue
from 0 to 10.34657359
from 0 to 0.50.11157178

Questions

What is the integral of x / (x2x^{2} + 1)?

The integral of xx2+1\frac{x}{x^{2} + 1} is ln⁡(x2+1)2+C\frac{\ln\left(x^{2} + 1\right)}{2} + C.

Why do we add + C?

Every function F(x) + C has the same derivative, so the antiderivative is only defined up to a constant.

What is the integral of x / (x2x^{2} + 1) from 0 to 1?

About 0.34657359.

More integrals

    Integral of x / (x^2 + 1): ln(x^2 + 1) / 2 + C (with steps) | WordGPT