∫arctg^3x/1+x^2 dx

интегрирование заменой переменной

To integrate the given function ∫arctg^3x/1+x^2 dx using variable substitution, we can let u = arctan(x).

Differentiating both sides with respect to x gives du/dx = 1/(1+x^2).

Rearranging the equation, we have dx = du/(1+x^2).

Substituting u and dx into the integral, we get ∫arctg^3x/1+x^2 dx = ∫u^3 * du/(1+x^2).

Now, we can rewrite the denominator as (1+u^2) using the substitution.

Therefore, the integral becomes ∫u^3 * du/(1+u^2).

To solve this integral, we can use the method of partial fractions.

First, we factorize the denominator as (1+u^2) = (u+i)(u-i).

Next, we can write the integrand as A/(u+i) + B/(u-i), where A and B are constants.

Multiplying through by (1+u^2), we get u^3 = A(u-i) + B(u+i).

Expanding and equating the coefficients of u^3, we have:

u^3 = (A+B)u - (Ai-Bi).

Comparing the coefficients, we get A + B = 0 and Ai - Bi = 0.

From the first equation, we have A = -B.

Substituting this into the second equation, we get Ai - Bi = 0, which implies A = B.

Therefore, A = -B.

Now, substituting these values back into the partial fraction decomposition, we have:

∫u^3 * du/(1+u^2) = ∫(-B/(u+i) + B/(u-i)) du.

Integrating each term separately, we get:

= -B*ln|u+i| + B*ln|u-i| + C,

where C is the constant of integration.

Finally, substituting back u = arctan(x), we get:

= -B*ln|arctan(x)+i| + B*ln|arctan(x)-i| + C.

Hence, the integral of arctan^3x/(1+x^2) dx using variable substitution is given by:

∫arctan^3x/(1+x^2) dx = -B*ln|arctan(x)+i| + B*ln|arctan(x)-i| + C.