Giải câu 2 bài: Nguyên hàm

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

Điều kiện: $x>0$.

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

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

<=> $\int f(x)dx=\frac3}{5} x^{\frac{5}{3}}+\frac{6}{7}\int x^{\frac{7}{6}}+\frac{3}{2} x^{\frac{2}{3}}+C$

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) $

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

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

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

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) $f(x)=\frac{1}{(1+x)(1-2x)}$

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

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

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

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