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

+) Ta có \(f\left( x \right) = \cos x.{\sin ^2}x = \cos x\left( {\frac{{1 - \cos 2x}}{2}} \right)\)\( = \frac{1}{2}\left( {\cos x - \cos x\cos 2x} \right)\)\( = \frac{1}{2}\cos x - \frac{1}{4}\left( {\cos 3x + \cos x} \right)\)\( = \frac{1}{4}\cos x - \frac{1}{4}\cos 3x\).

+) \(F\left( x \right) = \int {f\left( x \right){\rm{d}}x}  = \int {\left( {\frac{1}{4}\cos x - \frac{1}{4}\cos 3x} \right){\rm{d}}x} \)\( = \frac{1}{4}\sin \,x - \frac{1}{{12}}\sin \,3x + C\).

Lại có: \[F\left( 0 \right) = 2025\] \[ \Rightarrow C = 2025\]\( \Rightarrow F\left( x \right) = \frac{1}{4}\sin \,x - \frac{1}{{12}}\sin \,3x + 2025\)\( \Rightarrow F\left( {\frac{\pi }{2}} \right) = \frac{1}{4}\sin \,\frac{\pi }{2} - \frac{1}{{12}}\sin \,\frac{{3\pi }}{2} + 2025\)\( = \frac{1}{4} + \frac{1}{{12}} + 2025 = \frac{{6076}}{3} \approx 2025\).

+) Ta có \(f\left( x \right) = x{\left( {2x - 1} \right)^{2025}}\)\( = \frac{1}{2}\left( {2x - 1 + 1} \right){\left( {2x - 1} \right)^{2025}}\)\( = \frac{1}{2}{\left( {2x - 1} \right)^{2026}} + \frac{1}{2}{\left( {2x - 1} \right)^{2025}}\)..

+) \(F\left( x \right) = \int {f\left( x \right){\rm{d}}x} \)\( = \int {\left( {\frac{1}{2}{{\left( {2x - 1} \right)}^{2026}} + \frac{1}{2}{{\left( {2x - 1} \right)}^{2025}}} \right){\rm{d}}x} \)\( = \frac{1}{4}\frac{{{{\left( {2x - 1} \right)}^{2027}}}}{{2027}} + \frac{1}{4}\frac{{{{\left( {2x - 1} \right)}^{2026}}}}{{2026}} + C\)\( = \frac{{{{\left( {2x - 1} \right)}^{2027}}}}{{8108}} + \frac{{{{\left( {2x - 1} \right)}^{2026}}}}{{8104}} + C\).

Lại có: \[F\left( {\frac{1}{2}} \right) = 1\] \[ \Rightarrow C = 1\]\( \Rightarrow F\left( x \right) = \frac{{{{\left( {2x - 1} \right)}^{2027}}}}{{8108}} + \frac{{{{\left( {2x - 1} \right)}^{2026}}}}{{8104}} + 1\)\( \Rightarrow F\left( 0 \right) = \frac{{{{\left( { - 1} \right)}^{2027}}}}{{8108}} + \frac{{{{\left( { - 1} \right)}^{2026}}}}{{8104}} + 1 \approx 1\).