* Kỹ thuật thêm bớt.

\( \int \frac{x}{(x+1)^3} \, dx = \int \frac{x+1 - 1}{(x+1)^3} \, dx = \int \frac{1}{(x+1)^2} \, dx - \int \frac{1}{(x+1)^3} \, dx \)

\( = -\frac{1}{x+1} + \frac{1}{2(x+1)^2} + C \).

Cách 2: Đặt \( t = x+1 \Rightarrow dt = dx \) và \( x = t - 1 \).

\( I = \int \frac{t-1}{t^3} \, dt = \int \left( \frac{1}{t^2} - \frac{1}{t^3} \right) \, dt  = -\frac{1}{t} + \frac{1}{2t^2} + C \).

\( \int \frac{x}{(x+1)^2} \, dx = \int \frac{x+1 - 1}{(x+1)^2} \, dx = \int \left( \frac{1}{x+1} - \frac{1}{(x+1)^2} \right) dx \)

\( = \ln|x+1| + \frac{1}{x+1} + C \).

page65

\( \int \frac{1}{x(x^4 + 1)} \, dx = \int \frac{1 + x^4 - x^4}{x(x^4 + 1)} \, dx \)

\( = \int \frac{1}{x} \, dx - \int \frac{x^3}{1 + x^4} \, dx \)

\( = \ln|x| - \frac{1}{4} \ln(1 + x^4) + C \)

page66

\( \int \frac{1}{x^2(1+x)} \, dx = \int \frac{1 + x - x}{x^2(1+x)} \, dx \)

\( = \int \frac{1}{x^2} \, dx - \int \frac{1}{x(1+x)} \, dx \)

\( = -\frac{1}{x} - \int \left( \frac{1}{x} - \frac{1}{x+1} \right) \, dx \)

\( = -\frac{1}{x} - \ln \left| \frac{x}{x+1} \right| + C \)

page67 

\( I = \int \frac{1 + x^2 - 2x^2}{x(1 + x^2)} \, dx = \int \frac{1}{x} \, dx - \int \frac{2x}{1 + x^2} \, dx \)

\( = \ln|x| - \ln|1 + x^2| + C = \ln \left| \frac{x}{1 + x^2} \right| + C \quad \Rightarrow \boxed{\text{A}} \)

page68

Gợi ý: \( \int \sqrt{ax + b} \, dx \) là tính được

\( \int \frac{1}{\sqrt{x+1} + \sqrt{x-1}} \, dx = \frac{1}{2} \int \left( \sqrt{x+1} - \sqrt{x-1} \right) \, dx \)

\( = \frac{1}{2} \left[ \frac{2}{3}(x+1)\sqrt{x+1} - \frac{2}{3}(x-1)\sqrt{x-1} \right] + C \).

page69