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Một số bài tập nguyên hàm

Ảnh của tanphu
tanphu gửi vào T6, 21/02/2020 - 7:39ch

Tính các nguyên hàm

  1. \(\displaystyle\int\dfrac{1}{5^x}\left(\dfrac{\left(\sqrt{5}\right)^{2x+4}}{3x-2}+4^{2x}\right)\mathrm{d}x\)
  2. \(\displaystyle\int\dfrac{1}{\cos x}\left(1-\sin^2x-\dfrac{1}{\cos x}\right)\mathrm{d}x\)
  3. \(\displaystyle\int\dfrac{\mathrm{d}x}{\sin^2x.\cos^2x}\)

Giải

Câu 1. \(\displaystyle\int\dfrac{1}{5^x}\left(\dfrac{\left(\sqrt{5}\right)^{2x+4}}{3x-2}+4^{2x}\right)\mathrm{d}x=\displaystyle\int\dfrac{1}{5^x}\left(\dfrac{5^{x+2}}{3x-2}+16^x\right)\mathrm{d}x\)
\(=\displaystyle\int \dfrac{5^2}{3x-2}+\left(\dfrac{16}{5}\right)^x\mathrm{d}x=25.\dfrac{1}{3}\ln|3x-2|+\dfrac{\left(\dfrac{16}{5}\right)^x}{\ln\dfrac{16}{5}}+C\)

Câu 2. \(\displaystyle\int\dfrac{1}{\cos x}\left(1-\sin^2x-\dfrac{1}{\cos x}\right)\mathrm{d}x=\displaystyle\int\dfrac{1}{\cos x}\left(\cos^2x-\dfrac{1}{\cos x}\right)\mathrm{d}x\)
\(=\displaystyle\int\left(\cos x-\dfrac{1}{\cos^2x}\right)\mathrm{d}x=\sin x-\tan x+C.\)

Câu 3. \(\displaystyle\int\dfrac{\mathrm{d}x}{\sin^2x.\cos^2x}=\displaystyle\int\dfrac{\mathrm{d}x}{(\sin x.\cos x)^2}\)
\(=\displaystyle\int\dfrac{\mathrm{d}x}{\left(\dfrac{1}{2}\sin 2x\right)^2}=\displaystyle\int\dfrac{\mathrm{d}x}{\dfrac{1}{4}\sin^22x}\)
\(=\displaystyle\int\dfrac{4\mathrm{d}x}{\sin^22x}=-4.\dfrac{1}{2}\cot 2x +C=-2\cot 2x+C.\)

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