integrate 1/x^2
Let's integrate: <math><mo>∫(</mo><mfrac><mrow><mn>1</mn></mrow><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac><mo>)dx</mo></math> For the <u>math expression</u>: <math><mfrac><mrow><mn>1</mn></mrow><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac></math>, we know the derivative formula: <math><mo>(</mo><msup><mrow><mi>a</mi></mrow><mrow><mi>n</mi></mrow></msup><mo>)'</mo><mo> = </mo><mi>n</mi><mo>·</mo><msup><mrow><mi>a</mi></mrow><mrow><mo>(</mo><mi>n</mi><mo> -</mo><mn>1</mn><mo>)</mo></mrow></msup></math>, so we try differentiating with <m>^(m+1)</m> Let's perform the differentiation: <math><mo>(</mo><msup><mrow><mi>x</mi></mrow><mrow><mo>(</mo><mo>-</mo><mn>2</mn><mo> + </mo><mn>1</mn><mo>)</mo></mrow></msup><mo>)'</mo></math> <math><mo>(</mo><msup><mrow><mi>x</mi></mrow><mrow><mn>-1</mn></mrow></msup><mo>)'</mo></math> We differentiate using: <m><math><mo>(</mo><msup><mrow><mi>f</mi></mrow><mrow><mi>n</mi></mrow></msup><mo>)'</mo><mo> = </mo><mi>n</mi><mo>·</mo><mo>(</mo><mi>f</mi><mo>)'</mo><mo>·</mo><msup><mrow><mi>f</mi></mrow><mrow><mo>(</mo><mi>n</mi><mo> -</mo><mn>1</mn><mo>)</mo></mrow></msup></math></m> (where <m>f</m> = <m><math><mi>x</mi></math></m>) <math><mn>-1</mn><mo>·</mo><msup><mrow><mi>x</mi></mrow><mrow><mn>-2</mn></mrow></msup></math> <math><mo>-</mo><msup><mrow><mi>x</mi></mrow><mrow><mn>-2</mn></mrow></msup></math> <math><mfrac><mrow><mn>1</mn></mrow><mrow><mo>-</mo><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac></math> <math><mo>-</mo><mfrac><mrow><mn>1</mn></mrow><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup></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><mn>-1</mn><mo>·</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mi>x</mi></mrow></mfrac></math> <math><mo>-</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mi>x</mi></mrow></mfrac></math> Indefinite integrals are defined up to an additive constant <m>C</m>, so this yields: <math><mo>-</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mi>x</mi></mrow></mfrac><mo> + </mo><mi>C</mi></math> ▷<b>Answer: <math><mo>-</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mi>x</mi></mrow></mfrac><mo> + </mo><mi>C</mi></math></b><span style='font-size:smaller'>, where <m>C</m> is a constant (processing: 13 steps, 6 ms)</span>
