Question

hehe em i have seen how wolframalhpa solve this this integration, but is it possible to solve it if i choose not to expand all this equation out?

¶ (3x^2+ 7x^3)^3 dx

Answer 1

No

Answer 2

so even those it is (ax+bx^2)^99, we have to expand them all??

Answer 3

You could use the chain rule here. The chain rule goes like this:
Suppose you have a 2 functions f(x) and g(x), and f(x) and g(x) is differential at any point c (f'(c) and g'(c) exists) Then:
\[(f o g)(c) = f \prime(g(c)) * g \prime(c)\]
In your example:
Let \[f(x) = x ^{99}\]
\[g(x) = ax + b ^{2}\]
by applying chain rule:
\[f(g(x) = (ax + b x^{2})^{99} = (ax + bx ^{2})^{98} * (a + 2bx)\]

Answer 4

Use of integration by parts

Answer 5

You would anyway have to do same no of steps

Answer 6

oh sorry it says integration: Yeah you could do IBP

Answer 7

how to do 'integration by part' ? and wat is IBP?

Answer 8

IBP is integration by parts.

Answer 9

How do u separate this equation?

(ax+bx^2)^99

Answer 10

separate them into ax and bx^2 then use the chain rule?

Answer 11

This is pretty difficult you know, but alas i found the answer.
Integration by parts goes like this:
\[\int\limits f(x) g'(x)\, dx = f(x) g(x) - \int\limits f'(x) g(x)\, dx\]
or simply,
\[ \int\limits u\, dv=uv-\int\limits v\, du\! \]
Now to answer your question, lets extract the x from the "inner equation"
\[(3x^{2} + 7x^{3})^{3} = x^{6}(3+7x)^{3}\]
Therfore,
\[\int\limits_{}^{}(3x^{2}+7x^{3})^{3}dx=\int\limits_{}^{}x^{6}(3+7x)^{3}dx\]
Using IBP,
Let \[dv = (3+7x)^{3}dx\]
\[v = \frac{(3 + 7x)^{4}}{28}\]
\[u = x^{6}\]\[du = 6x^{5}dx\]
Solving the equation above:
\[∫x^{6}(3+7x)^{3}dx = \frac{(3+7x)^{4}}{28}* x^{6} - \int\limits_{}^{}\frac{(3+7x)^{4}}{28}*6x^{5}dx\]
Notice that we still have a integration to solve at the right side. I'll tackle that on next post.

Answer 12

So far we have solved the u * v part of the IBP but we are far from over. We still have an integration at the left. Again, lets do IBP, Let:
\[dv _{1} = (3 + 7x)^{4}dx\]\[v _{1} = \frac{(3 + 7x)^{5}}{35}\]\[u _{1} = \frac{6}{28}x^{5}\]\[du _{1} = \frac{30}{28}x^{4}dx\]
Solving again from IBP:
\[−∫\frac{(3+7x)^{4}}{28}∗6x^{5}dx = - (\frac{(3+7x)^{5}}{35}*\frac{6}{28}x^{5} - \int\limits_{}^{}\frac{(3+7x)^{5}}{35}*\frac{30}{28}x^{4}dx\]
Attaching the earlier result:
\[∫x^{6}(3+7x)^{3}dx=\frac{(3+7x)^{4}}{28}∗x^{6}−\frac{(3+7x)^{5}}{35}∗\frac{6}{28}x^{5}+∫\frac{(3+7x)^{5}}{35}∗\frac{30}{28}x^{4}dx\]
Notice the pattern here? If we repeat the IBP until the power of x is 0, we would have terms with powers of (3 + 7x) and x raise to something. Also the denominator = 7 * 7 * 5 * 4 or \[7^{2} * \frac{5!}{3!}\] the sign also alternates. So if we continue the trend:
\[∫x^{6}(3+7x)^{3}dx = \frac{3!6!(3+7x)^{4}}{7*4!*6!}∗x^{6}−\frac{3!6!(3+7x)^{5}}{7^{2}*5!*5!}∗x^{5}+\frac{3!6!(3+7x)^{6}}{7^{3}*6!*4!}∗x^{4} - \]
\[\frac{3!6!(3+7x)^{7}}{7^{4}*7!*3!}∗x^{3}+\frac{3!6!(3+7x)^{8}}{7^{5}*8!*2!}∗x^{2}-\frac{3!6!(3+7x)^{9}}{7^{6}*9!*2!}∗x^{2} + \frac{3!6!(3+7x)^{10}}{7^{7}*10!*1!}∗x^{1} - \]
\[\frac{3!6!(3+7x)^{11}}{7^{8}*11!*0!}\]

Answer 13

wow thats a long post XD