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Exercise 11.3 · Q5

Q.Integrate the following with respect to xx:
[!FORMULA] 61+(3x+2)2−121−(3−4x)2\dfrac{6}{1+(3x+2)^{2}}-\dfrac{12}{\sqrt{1-(3-4x)^{2}}}

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Both denominators are of the form 1+(ax+b)21+(ax+b)^2 and 1−(ax+b)21-(ax+b)^2 with a genuine linear (not purely scaled) argument; apply ∫dx1+(ax+b)2=1atan⁡−1(ax+b)+c\int\frac{dx}{1+(ax+b)^2}=\frac1a\tan^{-1}(ax+b)+c and ∫dx1−(ax+b)2=1asin⁡−1(ax+b)+c\int\frac{dx}{\sqrt{1-(ax+b)^2}}=\frac1a\sin^{-1}(ax+b)+c.

Step 1. First term. Here ax+b=3x+2ax+b=3x+2, so a=3a=3:

∫61+(3x+2)2 dx=6×13tan⁡−1(3x+2)+c=2tan⁡−1(3x+2)+c.\int \frac{6}{1+(3x+2)^2}\,dx = 6\times\frac13\tan^{-1}(3x+2)+c=2\tan^{-1}(3x+2)+c.

Step 2. Second term. Here ax+b=3−4x=−4x+3ax+b=3-4x=-4x+3, so a=−4a=-4:

∫121−(3−4x)2 dx=12×1−4sin⁡−1(3−4x)+c=−3sin⁡−1(3−4x)+c,\int \frac{12}{\sqrt{1-(3-4x)^2}}\,dx = 12\times\frac{1}{-4}\sin^{-1}(3-4x)+c=-3\sin^{-1}(3-4x)+c,

and since this term is subtracted in the original expression,

−∫121−(3−4x)2 dx=3sin⁡−1(3−4x)+c.-\int \frac{12}{\sqrt{1-(3-4x)^2}}\,dx = 3\sin^{-1}(3-4x)+c.

Step 3. Combine.

2tan⁡−1(3x+2)+3sin⁡−1(3−4x)+c.2\tan^{-1}(3x+2)+3\sin^{-1}(3-4x)+c. …

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