Evaluate the following indefinite integrals.

Show all steps please

A) ∫xe^x^2+10 dx
b) ∫(1/2x+x)(ln(x)+x^2)^10 dx

done yesterday

http://www.jiskha.com/display.cgi?id=1363367296

the first one is on the link, but not the second one

hint:

let's differentiate ( lnx + x^2)^11

we would get
11 (lnx + x^2)^10 ( 1/x + 2x)
= 11( lnx + x^2)^10 (2)( 1/(2x) + x)

= (11/2) ( 1/(2x) + x) (lnx + x^2)^10

mmmmhhh?

A) To evaluate the integral ∫xe^(x^2+10) dx, we can use the technique of substitution.

Let u = x^2 + 10. Differentiating both sides with respect to x gives du/dx = 2x. Solving for dx, we get dx = du / (2x).

Substituting these into the integral, we have ∫xe^(x^2+10) dx = ∫(1/2x)(e^u) du.

Now, we can split the integral into two parts:
∫(1/2x)e^u du = ∫(1/2) * (1/x) * e^u du.

The first part, ∫(1/2) du, is simply (1/2)u = (1/2)(x^2 + 10).

For the second part, we have ∫(1/x) * e^u du. Notice that this integral is the derivative of ln(x) with respect to u.

∫(1/x) * e^u du = ∫1 * e^u du = e^u = e^(x^2+10).

Putting it all together, we get:

∫xe^(x^2+10) dx = (1/2)(x^2 + 10) + e^(x^2+10) + C,

where C is the constant of integration.

B) To evaluate the integral ∫(1/2x+x)(ln(x)+x^2)^10 dx, we can expand the expression inside the parentheses and then use the power rule for integration.

Expanding (ln(x)+x^2)^10, we get:

(ln(x)+x^2)^10 = ∑[(10 choose k)(ln(x))^k(x^2)^(10-k)].

Using the power rule for integration, we can integrate each term separately:

∫(ln(x))^k(x^2)^(10-k) dx = ∫(ln(x))^k * x^(20-2k) dx.

The integral ∫(ln(x))^k dx can be evaluated using integration by parts with u = (ln(x))^k and dv = dx.

Differentiating u and integrating dv, we have:

du/dx = k(ln(x))^(k-1)(1/x),
v = x^(21-2k)/(21-2k).

Using the formula for integration by parts, ∫u dv = uv - ∫v du, we get:

∫(ln(x))^k * x^(20-2k) dx = (ln(x))^k * x^(21-2k)/(21-2k) - ∫k* (ln(x))^(k-1) * x^(21-2k)/(21-2k) * (1/x) dx.

Simplifying this expression, we have:

∫(ln(x))^k * x^(20-2k) dx = (ln(x))^k * x^(21-2k)/(21-2k) - k/(21-2k) ∫(ln(x))^(k-1) * x^(19-2k) dx.

Now, we can use this recursive formula to evaluate the integral in parts until we reach ∫(ln(x))^0 * x^(19-2(0)) dx = ∫x^19 dx, which can be directly integrated using the power rule.

After evaluating all the integrals, we can sum up the resulting terms and obtain the final indefinite integral.