Tìm nguyên hàm I = x^2 + 45 / x^4 -90 x^2 + 2025 dx

12/22

Tìm nguyên hàm \(I = \int {\frac{{{x^2} + 45}}{{{x^4} - 90{x^2} + 2025}}} {\rm{d}}x\).

\(I = \ln \left| {\frac{{{x^2} + x - 45}}{{{x^2} - x - 45}}} \right| + C\).

\(I = \ln \left| {\frac{{{x^2} - x - 45}}{{{x^2} + x - 45}}} \right| + C\).

\(I = \frac{1}{2}\ln \left| {\frac{{{x^2} + x - 45}}{{{x^2} - x - 45}}} \right| + C\).

\(I = \frac{1}{2}\ln \left| {\frac{{{x^2} - x - 45}}{{{x^2} + x - 45}}} \right| + C\).

Giải thích

Ta có: \(I = \int {\frac{{{x^2} + 45}}{{{x^4} - 91{x^2} + 2025}}} {\rm{d}}x\)\( = \int {\frac{{1 + \frac{{45}}{{{x^2}}}}}{{{x^2} - 91 + \frac{{2025}}{{{x^2}}}}}} {\rm{d}}x\)\( = \int {\frac{{\left( {1 + \frac{{45}}{{{x^2}}}} \right)}}{{{x^2} + \frac{{{{45}^2}}}{{{x^2}}} - 91}}} {\rm{d}}x\)\( = \int {\frac{{\left( {1 + \frac{{45}}{{{x^2}}}} \right)}}{{{{\left( {x - \frac{{45}}{x}} \right)}^2} - 1}}} {\rm{d}}x\)

\( = \int {\frac{1}{{{{\left( {x - \frac{{45}}{x}} \right)}^2} - 1}}} {\rm{d}}\left( {x - \frac{{45}}{x}} \right)\)\[ = \frac{1}{2}\left[ {\ln \left| {x - \frac{{45}}{x} - 1} \right| - \ln \left| {x - \frac{{45}}{x} + 1} \right|} \right]\]\( = \frac{1}{2}\ln \left| {\frac{{{x^2} - x - 45}}{{{x^2} + x - 45}}} \right| + C\).

Vậy \(I = \frac{1}{2}\ln \left| {\frac{{{x^2} - x - 45}}{{{x^2} + x - 45}}} \right| + C\).