Can trigonometric substitution be used to solve this integral? The integral in question is ∫x+19x2+6x+5dx I first completed the square in the denominator giving (3x+1)2+4 and proceeded to perform a u-substitution with u=3x+1, du=3 dx, and x=u−13. After simplifying, I was left with 19∫u+2u2+4du It is at this point I used trigonometric substitution with u=2tan⁡θ and du=2sec⁡2θdθ (I'm aware the integral can be written as 19∫uu2+4du+19∫2u2+4du and solved this way). After performing the trig substitution, I was left with 19∫(tan⁡θ+1)dθ=19ln|sec⁡θ|+19θ+C 19ln⁡((3x+1)2+42)+19tan−1(123x+1)+C which is incorrect. The correct answer is 118ln⁡(9x2+6x+5)+19tan−1(123x+1)+C What went wrong with my trig substitution?

Question

Can trigonometric substitution be used to solve this integral? The integral in question is ∫x+19x2+6x+5dx I first completed the square in the denominator giving (3x+1)2+4 and proceeded to perform a u-substitution with u=3x+1, du=3 dx, and x=u−13. After simplifying, I was left with 19∫u+2u2+4du It is at this point I used trigonometric substitution with u=2tan⁡θ and du=2sec⁡2θdθ (I&#039;m aware the integral can be written as 19∫uu2+4du+19∫2u2+4du and solved this way). After performing the trig substitution, I was left with 19∫(tan⁡θ+1)dθ=19ln|sec⁡θ|+19θ+C 19ln⁡((3x+1)2+42)+19tan−1(123x+1)+C which is incorrect. The correct answer is 118ln⁡(9x2+6x+5)+19tan−1(123x+1)+C What went wrong with my trig substitution?

Paineow · Accepted Answer

Since u=2tan⁡θ, you have  u2=4tan2θ=4(sec2θ−1)   So,   sec⁡θ=u2+44   and therefore    19ln⁡(sec⁡θ)=118ln⁡(u2+44)

alexandrebaud43 · Accepted Answer

There is a simpler way to determine this integral: rewrite first the numerator:  x+1=118(18x+6)+23   split the integral in two:   ∫x+19x2+6x+5dx=118∫18x+69x2+6x+5dx+23∫dx(3x+1)2+4  =118ln⁡(9x2+6x+5)+23∫dx(3x+1)2+4  =118ln⁡(9x2+6x+5)+29∫d(3x+1)(3x+1)2+4  and use the standard formula  ∫dxx2+a2=1aarctan⁡(xa)

nick1337 · Accepted Answer

∫x+19x2+6x+5dx=∫x+1(3x+1)2+22dx(u=3x+1)=19∫u+2u2+22du(u=2tan⁡θ)=19∫(2tan⁡θ+2)(2sec2⁡θ)22(1+tan2⁡θ)dθ=19∫(tan⁡θ+1)dθ=19ln⁡|sec⁡θ|+19θ+C0Now, |x|=x2 ⇒ln⁡|sec⁡θ|=ln⁡(sec2⁡θ)=12ln⁡sec2⁡θ=12ln⁡(1+tan2⁡θ) Which for the above: 19ln⁡|sec⁡θ|+19θ+C0=19×12ln⁡(1+tan2⁡θ)+19θ+C0=19×12ln⁡(1+(3x+12)2)+tan−1⁡(3x+12)+C0=118ln⁡(9x2+6x+5)+tan−1⁡(3x+12)+(C0−118ln⁡(4))=118ln⁡(9x2+6x+5)+tan−1⁡(3x+12)+C

Can trigonometric substitution be used to solve this integra

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