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If the solve the problem

Question:

$\int \tan ^{2}(2 x-3) d x$

Solution:

Let $I=\int \tan ^{2}(2 x-3) d x$

$=\int \tan ^{2}(2 x-3) d x$

$=\int \sec ^{2}(2 x-3)-1 d x$

Let $2 x-3=t d x=d t / 2$

$=\frac{1}{2} \int \sec ^{2} t-1 d t$

$=\frac{1}{2} \tan t-x$

Substitute the value of $t$

Hence, $I=\frac{1}{2} \tan (2 x-3)-x+C$

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