Sử dụng phương pháp đổi biến số để tính Nguyên hàm

Tìm họ các nguyên hàm sau

a) \(\int \frac{\ln ^{2} x-1}{x \ln x} \mathrm{~d} x\)

b) \(\int \frac{x \ln \left(x^{2}+1\right)}{x^{2}+1} \mathrm{~d} x\)

c) \(\int \frac{\ln ^{2} x}{x(1+\sqrt{\ln x+1})} d x\)

giải:

a) xét \(\int \frac{\ln ^{2} x-1}{x \ln x} \mathrm{~d} x\).

đặt t=lnx => dt \(=\frac{1}{x} \mathrm{~d} x\)

khi đó \(\int \frac{\ln ^{2} x-1}{x \ln x} \mathrm{~d} x=\int \frac{t^{2}-1}{t} \mathrm{~d} t\)

\(=\int\left(t-\frac{1}{t}\right) \mathrm{d} t=\frac{t^{2}}{2}-\ln |t|+C\)

\(=\frac{\ln ^{2} x}{2}-\ln |\ln x|+C\)

b) xét \(\int \frac{x \ln \left(x^{2}+1\right)}{x^{2}+1} \mathrm{~d} x\).

đặt t= \(\ln \left(x^{2}+1\right)\)

=> dt= \(\frac{2 x}{x^{2}+1} \mathrm{~d} x \Rightarrow \frac{1}{2} \mathrm{~d} t=\frac{x}{x^{2}+1} \mathrm{~d} x\).

c) xét \(\int \frac{\ln ^{2} x}{x(1+\sqrt{\ln x+1})} \mathrm{d} x\).

đặt t=1+\(\sqrt{1+\ln x}\)

=> \((t-1)^{2}=1+\ln x\)

<=> \(\ln x=t^{2}-2 t\)

suy ra \(\frac{\mathrm{d} x}{x}=(2 t-2) \mathrm{d} t\).

khi đó \(\int \frac{\ln ^{2} x}{x(1+\sqrt{\ln x+1})} \mathrm{d} x=\int \frac{\left(t^{2}-2 t\right)^{2}}{t} \cdot(2 t-2) \mathrm{d} t\)

\(=2 \int\left(t^{4}-5 t^{3}+8 t^{2}-4 t\right) \mathrm{d} t\)

\(=\frac{2}{5} t^{5}-\frac{5}{2} t^{4}+\frac{16}{3} t^{3}-4 t^{2}+C\).

như vậy

\(\int \frac{\ln ^{2} x}{x(1+\sqrt{\ln x+1})} \mathrm{d} x=\frac{2}{5} t^{5}-\frac{5}{2} t^{4}+\frac{16}{3} t^{3}-4 t^{2}+C\)

với  \(t=1+\sqrt{\ln x+1}\)

Tìm nguyên hàm 

a) \(I=\int(x+1) \sqrt[3]{3-2 x} d x\)

b) \(J=\int \frac{x d x}{\sqrt[3]{2 x+2}}\)

c) \(K=\int \frac{x d x}{\sqrt{x+3}+\sqrt{5 x+3}}\)

giải:

a) đặt \(t=\sqrt[3]{3-2 x} \Rightarrow x=\frac{3-t^{3}}{2} \Rightarrow d x=-\frac{3}{2} t^{2} d t\)

=> I=\(-\frac{3}{2} \int\left(\frac{3-t^{3}}{2}+1\right) t \cdot t^{2} d t=-\frac{3}{4} \int\left(5 t^{3}-t^{6}\right) d t\)

\(=-\frac{3}{4}\left(\frac{5 t^{4}}{4}-\frac{t^{7}}{7}\right)+C=\frac{3}{4}\left(\frac{\sqrt[3]{(3-2 x)^{7}}}{7}-\frac{5 \sqrt[3]{(3-2 x)^{4}}}{4}\right)+C\)

b) đặt \(t=\sqrt[3]{2 x+2} \Rightarrow x=\frac{t^{3}-2}{2} \Rightarrow d x=\frac{3}{2} t^{2} d t\)

suy ra 

\(J=\int \frac{\frac{t^{3}-2}{2} \frac{3}{2} t^{2} d t}{t}=\frac{3}{4} \int\left(t^{4}-2 t\right) d t=\frac{3}{4}\left(\frac{t^{5}}{5}-t^{2}\right)+C\)

\(=\frac{3}{4}\left(\frac{\sqrt[3]{(2 x+2)^{5}}}{5}-\sqrt[3]{(2 x+2)^{2}}\right)+C\).

c) \(I=\int \frac{x(\sqrt{5 x+3}-\sqrt{x+3}) d x}{5 x+3-x-3}\)

\(=\frac{1}{4} \int(\sqrt{5 x+3}-\sqrt{x+3}) d x\)

\(=\frac{1}{6}\left(\frac{1}{5} \sqrt{(5 x+3)^{3}}-\sqrt{(x+3)^{3}}\right)+C\).

Tìm nguyên hàm của

a) \(I=\int \sin ^{3} x \cdot \cos ^{5} x d x\)

b) \(J=\int \frac{\cos x d x}{(\sin x+2 \cos x)^{3}}\)

giải:

a) đặt t=cosx => dt=-sinxdc

ta có:

 \(I=\int\left(1-\cos ^{2} x\right) \cos ^{5} x \sin x d x=-\int\left(1-t^{2}\right) t^{5} d t\)

\(=\int\left(t^{7}-t^{5}\right) d t=\frac{t^{8}}{8}-\frac{t^{6}}{6}+C\)

\(=\frac{\sin ^{8} x}{8}-\frac{\sin ^{6} x}{6}+C\).

b) \(I=\int \frac{\cos x d x}{\cos ^{3} x(\tan x+2)^{3}}\)

\(=\int \frac{d x}{\cos ^{2} x(\tan x+2)^{3}}\)

đặt t=tanx => dt=\(\frac{1}{\cos ^{2} x} d x\)

do đó \(J=-\frac{1}{2} \frac{1}{(\tan x+2)^{2}}+C\)