Giải câu 2 bài: Nguyên hàm

Giải câu 2 bài: Nguyên hàm.

a)  $f(x)=\frac{x+\sqrt{x}+1}{\sqrt[3]{x}}$

Điều kiện: $x>0$.

<=> $f(x)=x^{\frac{2}{3}}+x^{\frac{1}{6}}+x^{\frac{1}{3}}$

=> $\int f(x)dx=\int x^{\frac{2}{3}}dx+\int x^{\frac{1}{6}}dx+\int x^{\frac{1}{3}}dx$

<=> $\text{Extra close brace or missing open brace}$

b) $f(x)=\frac{2^x-1}{e^x}$

=> $\int f(x)dx=\int \frac{2^x-1}{e^x}dx$

<=> $\int f(x)dx=\int (\frac{2}{e}{)}^{x}dx-\int e^{-x}dx$

<=> $\int f(x)dx=\frac{2^x}{e^x(\ln 2-1)}+\frac{1}{e^x}+C$

c) $f(x)=\frac{1}{{\sin }^{2}x.{\cos }^{2}x}$

<=> $f(x)=\frac{4}{{\sin }^{2}2x}$

=> $\int f(x)dx=\int \frac{4}{{\sin }^{2}2x}dx=-2\cot 2x+C$ 

d) $f(x)=\sin 5x.\cos 3x$

<=> $f(x)=\frac{1}{2}(\sin 8x+\sin 2x)$

=> $\int f(x)dx=\int \frac{1}{2}(\sin 8x+\sin 2x)dx$

<=> $\int f(x)dx=\frac{1}{2}\int \sin 8xdx+\frac{1}{2}\int \sin 2xdx$

<=> $\int f(x)dx=-\frac{1}{16}\cos 8x-\frac{1}{4}\cos 2x+C$

e) $f(x)={\tan }^{2}x$

<=> $f(x)=\frac{1}{{\cos }^{2}x}-1$

=> $\int f(x)dx=\int (\frac{1}{{\cos }^{2}x}-1)dx$

<=> $\int f(x)dx=\tan x-x+C$

g) $f(x)=e^{3-2x}$

=> $\int f(x)dx=\int e^{3-2x}dx$

<=> $\int f(x)dx=\frac{-1}{2}{e}^{3-2x}+C$

h) $f(x)=\frac{1}{(1+x)(1-2x)}$

<=> $f(x)=\frac{1}{(3(1+x))}+\frac{2}{(3(1-2x))}$

=> $\int f(x)dx=\int (\frac{1}{3(1+x)}+\frac{2}{3(1-2x)})dx$

<=> $\int f(x)dx=\frac{1}{3}\ln\left | 1+x \right |-\frac{1}{3\ln\left | 1-2x \right |$

<=> $\int f(x)dx=\frac{1}{3}\left|\frac{1+x}{1-2x}\right|+C$