Nguyên hàm [Lần 1]

Bài 1. Tìm nguyên hàm: $\int x\left(1-\dfrac{e^x}{x} \right)dx$ 

$$\begin{aligned} \int x(1- \frac{e^x}{x}) \text{dx}  &  = \int x \text{dx} - \int e^x \text{dx}  \\ & = \frac{x^2}{2} - e^x + C\end{aligned}$$ [*]$$

Bài 2. Tìm nguyên hàm $\int\dfrac{x^2-x+3}{x+1}dx$

$$\begin{aligned} \int \frac{x^2-x+3}{x+1} \text{dx}  &  = \int (x-2) \text{dx} + \int \frac{5}{x+1} \text{dx}  \\ & = \frac{(x-2)^2}{2} + 5 \ln | x+1| +C \end{aligned}$$

Bài 3. Tìm nguyên hàm: $$\int \dfrac{x^4+1}{x+1}dx$$

Ta có: $\dfrac{x^4+1}{x+1}=x^3-x^2+x-1+\dfrac{2}{x+1}$
Do đó: $$\int \dfrac{x^4+1}{x+1}dx=\int \left(x^3-x^2+x-1+\dfrac{2}{x+1}\right)dx$$
$$=\dfrac{x^4}{4}-\dfrac{x^3}{3}+\dfrac{x^2}{2}+2\ln |x+1|+C$$

Bài 4.  $$\int \dfrac{dx}{\sqrt{(x^2+1)^3}}$$

$\sqrt{x^2+1}$ thì đặt $x = \tan  t$ và sau khi đặt thì ta được:
$$\displaystyle I = \int \frac{\frac{1}{\cos^2 t}}{\frac{1}{\cos ^3t}} \text{dt } = \int \text{d ( sin t)} = \sin t = \frac{\tan t}{\sqrt{1+ \tan^2 t}} = \frac{x}{\sqrt{x^2+1}}$$
$t = \frac{x}{\sqrt{x^2+1}}$
Đặt $\displaystyle t = \frac{x}{\sqrt{x^2+1}} \Rightarrow dt =\frac{\sqrt{x^2+1}-\frac{x^2}{\sqrt{x^2+1}}}{x^2+1} \text{dx } = \frac{1}{(\sqrt{x^2+1})^3} \text{dx }$
Và ta có:
$$I = \int \text{dt} = t = \frac{x}{\sqrt{x^2+1}}$$
$\displaystyle\int \dfrac{\mbox{d}x}{\sqrt{(x^2+1)^3}}=\int \frac{\frac{1}{\sqrt{x^2+1}}}{x^2+1} \mbox{d}x=\int \frac{\sqrt{x^2+1}-\frac{x^2}{\sqrt{x^2+1}}}{x^2+1} \mbox{d}x= \int \mbox{d}\left(\frac{x}{\sqrt{x^2+1}}\right)$
$=\frac{x}{\sqrt{x^2+1}}+C$

Bài 5. Tìm nguyên hàm: $\int \sqrt[4]{x\sqrt[3]{x}} dx$

Ta có $\displaystyle \int \sqrt[4]{x \sqrt[3]{x}}\mbox{d}x =\displaystyle \int x^{\frac{1}{3}}\mbox{d}x$
$= \dfrac{3}{4}x \sqrt[3]{x} +C$