MHB Integrating xarctg2x Using Integration by Parts

[leprofece]

it is integral of xarctg²x
u = arctg²x
du =1/(1+x²) or 2(arctgx/(1+x²) ?' I am stuck here

dv = x v =x²/2

 

[MarkFL]

I would consider:

$$u=\tan^{-1}(x)$$

$$dv=x\tan^{-1}(x)$$

Finding $du$ is easy, but you will want to use integration by parts to find $v$.

 

[leprofece]

I would consider:

$$u=\tan^{-1}(x)$$

$$dv=x\tan^{-1}(x)$$

Finding $du$ is easy, but you will want to use integration by parts to find $v$.

Ok as it is to the square
du = 2arctgx/(1+x²)
and dv = x consequently v = x²/2

so applying formula I got (tang^-1(x))²(x²)/2 - 1/2integral (x² (arctgx))/(1+x²)

Now how can I solve this second integral??

 

[Opalg]

Ok as it is to the square
du = 2arctgx/(1+x²)
and dv = x consequently v = x²/2

so applying formula I got (tang^-1(x))²(x²)/2 - 1/2integral (x² (arctgx))/(1+x²)

Now how can I solve this second integral??

To evaluate $$\int \frac{x^2 \arctan x}{1+x^2}dx$$, write $$\frac{x^2}{1+x^2} = \frac{(1+x^2)-1}{1+x^2} = 1 - \frac1{1+x^2},$$ so that $$\int \frac{x^2 \arctan x}{1+x^2}dx = \int\arctan x\,dx - \int \frac{\arctan x}{1+x^2}dx.$$ Then see if you can evaluate each of those last two integrals (using integration by parts again).

 

[leprofece]

That is the end
= {(tan⁻¹x)²}*(x²/2) - ∫{(tan⁻¹x)/(1 + x²)}*(x²) dx

= {(tan⁻¹x)²}*(x²/2) - ∫{(tan⁻¹x) dx + ∫{(tan⁻¹x)dx/(1 + x²) --------- (1)

iii) ∫{(tan⁻¹x)dx/(1 + x²); let (tan⁻¹x) = y; so, dx/(1 + x²) = dy
==> ∫{(tan⁻¹x)dx/(1 + x²) = ∫y dy = y²/2 = (tan⁻¹x)²/2 -------- (A)

∫{(tan⁻¹x) dx = x*(tan⁻¹x) - ∫x* dx/(1 + x²)
∫x* dx/(1 + x²) = (1/2)∫2x* dx/(1 + x²) = (1/2)*ln|1 + x²|
So, ∫{(tan⁻¹x) dx = x*(tan⁻¹x) - (1/2)*ln|1 + x²| -------------- (B)

Substituting the values as in (A) & (B) in (1),
∫{(tan⁻¹x)/(1 + x²)}*(x²) dx = (tan⁻¹x)²/2 - x*(tan⁻¹x) + (1/2)*ln|1 + x²| ------ (3)

∫x*{(tan⁻¹x)²} dx = {(tan⁻¹x)²}*(x²/2) + (tan⁻¹x)²/2 - x*(tan⁻¹x) + (1/2)*ln|1 + x²| + C

 

[Prove It]

To evaluate $$\int \frac{x^2 \arctan x}{1+x^2}dx$$, write $$\frac{x^2}{1+x^2} = \frac{(1+x^2)-1}{1+x^2} = 1 - \frac1{1+x^2},$$ so that $$\int \frac{x^2 \arctan x}{1+x^2}dx = \int\arctan x\,dx - \int \frac{\arctan x}{1+x^2}dx.$$ Then see if you can evaluate each of those last two integrals (using integration by parts again).

Except the integral is $\displaystyle \begin{align*} \int{ \frac{x\arctan^2{(x)}}{1 + x^2}\,dx} \end{align*}$, not $\displaystyle \begin{align*} \int{ \frac{x^2\arctan{(x)}}{1 + x^2}\,dx} \end{align*}$.

 

[Opalg]

Except the integral is $\displaystyle \begin{align*} \int{ \frac{x\arctan^2{(x)}}{1 + x^2}\,dx} \end{align*}$, not $\displaystyle \begin{align*} \int{ \frac{x^2\arctan{(x)}}{1 + x^2}\,dx} \end{align*}$.

The second of those integrals is the one that occurs in comment #3 above.