Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
Tuyệt vời, đợi mình load rồi mình hỏi thêm vào câu nữa nha bẹn
![](https://rs.olm.vn/images/avt/0.png?1311)
a. \(\int\dfrac{x^3}{x-2}dx=\int\left(x^2+2x+4+\dfrac{8}{x-2}\right)dx=\dfrac{1}{3}x^3+x^2+4x+8ln\left|x-2\right|+C\)
b. \(\int\dfrac{dx}{x\sqrt{x^2+1}}=\int\dfrac{xdx}{x^2\sqrt{x^2+1}}\)
Đặt \(\sqrt{x^2+1}=u\Rightarrow x^2=u^2-1\Rightarrow xdx=udu\)
\(I=\int\dfrac{udu}{\left(u^2-1\right)u}=\int\dfrac{du}{u^2-1}=\dfrac{1}{2}\int\left(\dfrac{1}{u-1}-\dfrac{1}{u+1}\right)du=\dfrac{1}{2}ln\left|\dfrac{u-1}{u+1}\right|+C\)
\(=\dfrac{1}{2}ln\left|\dfrac{\sqrt{x^2+1}-1}{\sqrt{x^2+1}+1}\right|+C\)
c. \(\int\left(\dfrac{5}{x}+\sqrt{x^3}\right)dx=\int\left(\dfrac{5}{x}+x^{\dfrac{3}{2}}\right)dx=5ln\left|x\right|+\dfrac{2}{5}\sqrt{x^5}+C\)
d. \(\int\dfrac{x\sqrt{x}+\sqrt{x}}{x^2}dx=\int\left(x^{-\dfrac{1}{2}}+x^{-\dfrac{3}{2}}\right)dx=2\sqrt{x}-\dfrac{1}{2\sqrt{x}}+C\)
e. \(\int\dfrac{dx}{\sqrt{1-x^2}}=arcsin\left(x\right)+C\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{d}{dx}\left(f\left(x\right)\right)\equiv f'\left(x\right)\)
\(\dfrac{1}{sinx}dx=\dfrac{sinx}{sin^2x}dx=\dfrac{sinx}{1-cos^2x}dx=\dfrac{d\left(cosx\right)}{cos^2x-1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\int\dfrac{dx}{x^3+x}=\int\dfrac{dx}{x\left(x^2+1\right)}\)
\(t=x^2+1\Rightarrow dt=2xdx\Rightarrow\int\dfrac{dx}{x\left(x^2+1\right)}=\int\dfrac{dt}{2x^2t}=\dfrac{1}{2}\int\dfrac{dt}{\left(t-1\right).t}\)
\(\dfrac{1}{\left(t-1\right).t}=\dfrac{1}{t-1}-\dfrac{1}{t}\)
\(\Rightarrow\int\dfrac{dt}{\left(t-1\right)t}=\int\left(\dfrac{1}{t-1}-\dfrac{1}{t}\right)dt=\int\dfrac{dt}{t-1}-\int\dfrac{dt}{t}=ln\left|t-1\right|-ln\left|t\right|=ln\left|x^2\right|-ln\left|x^2+1\right|\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\int sin^2\dfrac{x}{2}dx=\int\left(\dfrac{1}{2}-\dfrac{1}{2}cosx\right)dx=\dfrac{1}{2}x-\dfrac{1}{2}sinx+C\)
\(\int cos^23xdx=\int\left(\dfrac{1}{2}+\dfrac{1}{2}cos6x\right)dx=\dfrac{1}{2}x+\dfrac{1}{12}sin6x+C\)
\(\int4cos^2\dfrac{x}{2}dx=\int\left(2+2cosx\right)dx=2x+2sinx+C\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Sử dụng nguyên hàm từng phần:
\(I=\int x.sin\dfrac{x}{3}dx\) \(\Rightarrow\) đặt \(\left\{{}\begin{matrix}u=x\\dv=sin\dfrac{x}{3}dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=dx\\x=-3cos\dfrac{x}{3}\end{matrix}\right.\)
\(\Rightarrow I=-3x.cos\dfrac{x}{3}+3\int cos\dfrac{x}{3}dx=-3x.cos\dfrac{x}{3}+9sin\dfrac{x}{3}+C\)
àấdgfd