ai giảng cho mk câu này vs ạ tính tích phân a) I= `\int_{1}^{0} x(1-x^2)^4dx` b) I= `\int_{√7}^{4}\frac{dx}{x\sqrt{x^2+9}} `

Đáp án:

$a)I=-\dfrac{1}{10}\\ b)I=\dfrac{1}{6} \ln \dfrac{7}{4}.$

Giải thích các bước giải:

$a)I=\displaystyle\int\limits_1^0 x(1-x^2)^4 \, dx\\ =\dfrac{1}{2}\displaystyle\int\limits_1^0 (x^2-1)^4 \, d(x^2-1)\\ =\dfrac{1}{2}.\dfrac{1}{5}(x^2-1)^5\Bigg\vert_1^0\\ =\dfrac{1}{10}(x^2-1)^5\Bigg\vert_1^0\\ =-\dfrac{1}{10}\\ b)I=\displaystyle\int\limits_\sqrt{7}^4 \dfrac{dx}{x\sqrt{x^2+9}}=\displaystyle\int\limits_\sqrt{7}^4 \dfrac{xdx}{x^2\sqrt{x^2+9}}\\ t=\sqrt{x^2+9} \Rightarrow t^2=x^2+9 \Rightarrow x^2=t^2-9 \Rightarrow xdx  =tdt\\ \begin{array}{|c|c|}\hline x&\sqrt{7}&4\\\hline t&4&5 \\\hline\end{array}\\ I=\displaystyle\int\limits_4^5 \dfrac{tdt}{(t^2-9)t}\\ =\displaystyle\int\limits_4^5 \dfrac{dt}{t^2-9}\\ =\displaystyle\int\limits_4^5 \dfrac{dt}{(t-3)(t+3)}\\ =\dfrac{1}{6}\displaystyle\int\limits_4^5 \left(\dfrac{1}{t-3}-\dfrac{1}{t+3}\right)\, dt\\ =\dfrac{1}{6} \ln \left|\dfrac{t-3}{t+3}\right| \Bigg\vert_4^5\\ =\dfrac{1}{6} \left(\ln \dfrac{1}{4}-\ln \dfrac{1}{7}\right)\\ =\dfrac{1}{6} \ln \dfrac{7}{4}.$