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Exercise 7.4 · Q5

Q.Integrate the following function: 3x1+2x4\frac{3x}{1+2x^4}

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The substitution u=x2u=x^2 turns this into an arctangent integral, giving 324tan⁡−1(2 x2)+C\frac{3\sqrt2}{4}\tan^{-1}(\sqrt2\,x^2)+C.

Choose the substitution

The denominator involves x4=(x2)2x^4=(x^2)^2 and the numerator carries a single xx — a textbook cue for u=x2u=x^2, because du=2x dxdu=2x\,dx absorbs that xx. Then x dx=du2x\,dx=\frac{du}{2} and

∫3x1+2x4 dx=∫31+2u2⋅du2=32∫du1+2u2.\int\frac{3x}{1+2x^4}\,dx=\int\frac{3}{1+2u^2}\cdot\frac{du}{2}=\frac32\int\frac{du}{1+2u^2}.

Match the arctangent form

Factor the 22 out of the denominator: 1+2u2=2(u2+12)1+2u^2=2\left(u^2+\frac12\right), so

32∫du2(u2+12)=34∫duu2+(12)2.\frac32\int\frac{du}{2\left(u^2+\frac12\right)}=\frac34\int\frac{du}{u^2+\left(\frac{1}{\sqrt2}\right)^2}.

With A=12A=\frac{1}{\sqrt2}, use ∫duu2+A2=1Atan⁡−1uA\int\frac{du}{u^2+A^2}=\frac1A\tan^{-1}\frac{u}{A}:

34⋅11/2tan⁡−1 ⁣(u1/2)+C=34⋅2 tan⁡−1(2 u)+C=324tan⁡−1(2 u)+C.\frac34\cdot\frac{1}{1/\sqrt2}\tan^{-1}\!\left(\frac{u}{1/\sqrt2}\right)+C=\frac34\cdot\sqrt2\,\tan^{-1}(\sqrt2\,u)+C=\frac{3\sqrt2}{4}\tan^{-1}(\sqrt2\,u)+C.

Back-substitute and check …

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