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