integre 1/(x+1)
Let's compute the integral: <math><mo>∫(</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mi>x</mi><mo> + </mo><mn>1</mn></mrow></mfrac><mo>)dx</mo></math> We attempt to reverse an integration: starting from <math><mfrac><mrow><mn>1</mn></mrow><mrow><mi>x</mi><mo> + </mo><mn>1</mn></mrow></mfrac></math> and <math><mo>(</mo><mi>ln</mi><mo>(</mo><mi>a</mi><mo>)</mo><mo>)'</mo><mo> = </mo><mfrac><mrow><mn>1</mn></mrow><mrow><mi>a</mi></mrow></mfrac></math>, we differentiate using <m>ln</m> Let's perform the differentiation: <math><mo>(</mo><mi>ln</mi><mo>(</mo><mi>x</mi><mo> + </mo><mn>1</mn><mo>)</mo><mo>)'</mo></math> We differentiate using: <m><math><mo>(</mo><mi>ln</mi><mo>(</mo><mi>f</mi><mo>)</mo><mo>)'</mo><mo> = </mo><mo>(</mo><mi>f</mi><mo>)'</mo><mo>·</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mi>f</mi></mrow></mfrac></math></m> (where <m>f</m> = <m><math><mi>x</mi><mo> + </mo><mn>1</mn></math></m>) <math><mo>(</mo><mi>x</mi><mo> + </mo><mn>1</mn><mo>)'</mo><mo>·</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mi>x</mi><mo> + </mo><mn>1</mn></mrow></mfrac></math> <math><mfrac><mrow><mn>1</mn><mo> + </mo><mn>0</mn></mrow><mrow><mi>x</mi><mo> + </mo><mn>1</mn></mrow></mfrac></math> <math><mfrac><mrow><mn>1</mn></mrow><mrow><mi>x</mi><mo> + </mo><mn>1</mn></mrow></mfrac></math> You observe that the computed derivative and the original <u>math expression</u> are the same, except for a coefficient. So the integral is: <math><mi>ln</mi><mo>(</mo><mi>x</mi><mo> + </mo><mn>1</mn><mo>)</mo></math> Indefinite integrals are defined up to an additive constant <m>C</m>, so this yields: <math><mi>ln</mi><mo>(</mo><mi>x</mi><mo> + </mo><mn>1</mn><mo>)</mo><mo> + </mo><mi>C</mi></math> ▷<b>Result: <math><mi>ln</mi><mo>(</mo><mi>x</mi><mo> + </mo><mn>1</mn><mo>)</mo><mo> + </mo><mi>C</mi></math></b><span style='font-size:smaller'>, where <m>C</m> is a constant (computation required 10 steps and 2 ms)</span>
