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Definite integralsCalculusCollegeabout 1:01

Integrate x/(x^2+1) from 0 to 2: animated lesson

A narrated, step-by-step definite integrals video lesson with captions. Final answer: 0.8047189562.

Video with a synthetic English voice and captions. Pick another language for subtitles in that language, or open the interactive version.

Video narrated by a synthetic voice (Piper TTS, en_US-ljspeech-high voice (LJ Speech dataset, public domain)); the steps come from our maths engine and the answer was checked independently. Download MP4 (0.4 MB) · Captions (WebVTT)

Steps on screen

  1. 1. The problem
    Evaluating an integral
    ∫02xx2+1 dx\int_{0}^{2} \frac{x}{{x}^{2}+1}\,dx
  2. 2. Step by step
    1. ∫02xx2+1 dx\int_{0}^{2} \frac{x}{{x}^{2}+1}\,dx
      Start with the definite integral
    2. ∫xx2+1 dx=ln⁡(∣x2+1∣)2\int \frac{x}{{x}^{2}+1}\,dx = \frac{\ln\left(\left|{x}^{2}+1\right|\right)}{2}
      Logarithm rule: the numerator is a constant multiple of the denominator's derivative, ∫ u′/u dx = ln|u|
    3. F(x)=ln⁡(∣x2+1∣)2F(x) = \frac{\ln\left(\left|{x}^{2}+1\right|\right)}{2}
      Antiderivative
    4. F(2)−F(0)=0.8047189562−0=0.8047189562F(2) - F(0) = 0.8047189562 - 0 = 0.8047189562
      Apply the fundamental theorem of calculus
  3. 3. The answer
    ∫02xx2+1 dx=0.8047189562\int_{0}^{2} \frac{x}{{x}^{2}+1}\,dx = 0.8047189562

    ∫ from 0 to 2 = 0.8047189562

  4. 4. Check
    • Differentiated the antiderivative and compared it with the integrand ✓
    • Adaptive Simpson's rule gives 0.8047189562 ✓

Transcript

The problem. Here is the problem: integrate x/(x squared plus 1) from 0 to 2. Evaluate the definite integral of x / (x squared plus 1) from x equals 0 to x equals 2.

Step by step. Start with the definite integral. Logarithm rule: the numerator is a constant multiple of the denominator prime s derivative, the integral of u prime /u dx equals ln|u|. Antiderivative. Apply the fundamental theorem of calculus.

The answer. So the answer is 0.8047189562. A definite integral gives the signed area between the curve and the x-axis: area above the axis counts as positive and area below counts as negative.

Check. Differentiated the antiderivative and compared it with the integrand, which checks out. Adaptive Simpson prime s rule gives 0.8047189562, which checks out.

About definite integrals

Evaluate definite integrals with the fundamental theorem of calculus.

Integrate x/(x^2+1) from 0 to 2: Definite integrals Lesson