How do you integrate ((1/5)x-(1/5))/(x^2 + 4)?

Bring the 1/5 in front of the integral for simplification.

1/5 ∫(x - 1)/(x² + 4)dx

You can now split this into a sum of two integrals.

1/5 ∫x/(x² + 4)dx - 1/5 ∫1/(x² + 4)dx

For the first integral, use a u substitution where u = x² + 4 -> du = 2xdx.

1/5 ∫x/(x² + 4)dx

= 1/10 ∫1/u du

= (1/10) ln(x² + 4) + C

For the second integral, you should recognize that 1/(x² + 4) will require a trigonometric substitution where x = 2tan(θ), dx = 2sec²(θ)dθ, θ = arctan(x/2)

1/5 ∫1/(x² + 4)dx

= 1/5 ∫2sec²(θ)/(4tan²(θ)+4) dθ

= 1/5 ∫2sec²(θ)/4(tan²(θ)+1) dθ

= 1/10 ∫sec²(θ)/(sec²(θ)) dθ

= 1/10 ∫ dθ

= (1/10) θ + C

= (1/10) arctan(x/2) + C

So we have the values for the two integrals, which we can use to find the solution to the original integral.

1/5 ∫x/(x² + 4)dx - 1/5 ∫1/(x² + 4)dx

= (1/10) ln(x² + 4) - (1/10) arctan(x/2) + C

= (1/10)[ln(x² + 4) - arctan(x/2)] + C is your solution.

= (1/10)ln(x^2+4) - (1/10)tan^-1 (x/2) + C

I am going to admit, I forgot the +c at the last part.

Sorry about the bad handwriting and head tilt.

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