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Câu hỏi: Họ nguyên hàm của hàm số $f\left( x \right)={{\tan }^{2}}x$ là
A. $\int{f\left( x \right)\text{d}x=\tan x+C}$.
B. $\int{f\left( x \right)\text{d}x=x-\tan x+C}$.
C. $\int{f\left( x \right)\text{d}x=\tan x+x+C}$.
D. $\int{f\left( x \right)\text{d}x=\tan x-x+C}$.
A. $\int{f\left( x \right)\text{d}x=\tan x+C}$.
B. $\int{f\left( x \right)\text{d}x=x-\tan x+C}$.
C. $\int{f\left( x \right)\text{d}x=\tan x+x+C}$.
D. $\int{f\left( x \right)\text{d}x=\tan x-x+C}$.
Ta có $1+{{\tan }^{2}}x=\dfrac{1}{{{\cos }^{2}}x}\Rightarrow {{\tan }^{2}}x=\dfrac{1}{{{\cos }^{2}}x}-1$.
Suy ra $\int{f(x)\text{d}x=\int{{{\tan }^{2}}x\text{d}x=\int{\left( \dfrac{1}{{{\cos }^{2}}x}-1 \right)\text{d}x=\tan x-x+C}}}$.
Suy ra $\int{f(x)\text{d}x=\int{{{\tan }^{2}}x\text{d}x=\int{\left( \dfrac{1}{{{\cos }^{2}}x}-1 \right)\text{d}x=\tan x-x+C}}}$.
Đáp án D.