Integral of x^2 arctan (4x)?

Hello,

let's write the integral as:

∫ arctan(4x) x² dx =

let:

arctan(4x) = u → (also 4, that is the derivative of the argument, is needed) →

4 {1 /[(4x)² + 1]} dx = [4 /(16x² + 1)] dx = du

x² dx = dv → [1/(2+1)] x² ⁺ ¹ = (1/3)x³ = v

then, integrating by parts:

∫ u dv = v u - ∫ v du

∫ arctan(4x) x² dx = (1/3)x³ arctan(4x) - ∫ (1/3)x³ [4 /(16x² + 1)] dx =

(factoring constants out)

(1/3)x³ arctan(4x) - (4/3) ∫ [x³ /(16x² + 1)] dx (#)

instead of long dividing, I think it more convenient to do as follows:

let's rewrite the remaining integral as:

∫ [x² /(16x² + 1)] x dx =

let:

16x² + 1 = t

16x² = t - 1

x² = (t - 1)/16

(differentiating both sides)

d(x²) = d[(t - 1)/16]

2x dx = (1/16) dt

x dx = (1/2)(1/16) dt = (1/32) dt

then, substituting:

∫ [x² /(16x² + 1)] x dx = ∫ {[(t - 1)/16] /t} (1/32) dt =

∫ (1/512) [(t - 1) /t] dt =

(distributing and simplifying)

∫ (1/512) [(t /t) - (1 /t)] dt =

∫ (1/512) [1 - (1 /t)] dt =

(splitting into two integrals and factoring constants out)

(1/512) ∫ dt - (1/512) ∫ (1 /t) dt =

(1/512)t - (1/512) ln | t | + C

let's substitute back 16x² + 1 for u:

(1/512)(16x² + 1) - (1/512) ln |16x² + 1| + C =

(being the argument of the logarithm always positive, the absolute value is not needed any longer)

(1/32)x² + (1/512) - (1/512) ln (16x² + 1) + C =

being 1/512 a constant, we can include it in C:

(1/32)x² - (1/512) ln (16x² + 1) + C

let's plug this into the above (#) expression, obtaining:

(1/3)x³ arctan(4x) - (4/3) [(1/32)x² - (1/512) ln (16x² + 1)] + C =

(1/3)x³ arctan(4x) - (1/24)x² + (1/384) ln (16x² + 1) + C (answer)

I hope it's helpful

You should have du = 4 / (16*x^2 + 1), so the end result should be

(1/3)*x^3 * arctan(4x) - (4/3)*integral((x^3 / (16*x^2 + 1)) dx).

Now by long division we get x^3 / (16*x^2 + 1) = (1/16)*(x - (x / (16*x^2 + 1))),

so the integral now becomes

(1/3)*x^3 * arctan(4x) - (1/12)*integral((x - (x / (16*x^2 + 1))) dx) =

(1/3)*x^3 * arctan(4x) - (1/24)*x^2 + (1/12)*integral((x / (16*x^2 + 1)) dx).

For this last integral, let w = 16*x^2 + 1; then dw = 32*x dx, which makes

(1/12)*integral((x / (16*x^2 + 1)) dx) =

(1/384)*integral((1/w) dw) = (1/384)*ln lwl + C = (1/384)*ln(16*x^2 + 1) + C,

where I could drop the absolute value signs around w since 16*x^2 + 1 > 0.

So the final answer is

(1/3)*x^3 * arctan(4x) - (1/24)*x^2 + (1/384)*ln(16*x^2 + 1) + C.

Integral Of X 2