Question

Integration of (3X(X^2+3)^4) dxlass="latex-formula">

Answer 1

Answer:

Step-by-step explanation:

$\int 3x(x^2+3)^4 \ dx$.

It is apparently obvious we could expand the bracket and integrate term-by-term. This method would work but is very time consuming (and you could easily make a mistake) so we use a different method: integration by substitution.

Integration by substitution involves swapping the variable $x$ for another variable which depends on x: $u(x)$. (We are going to choose $u$ for this question).

The very first step is to choose a suitable substitution. That is, an equation $u=f(x)$ which is going to make the integration easier. There is a trick for spotting this however: if an integral contains both a term and it's derivative then use the substitution $u=\text{The Term}$.

Your integral contains the term $x^2 + 3$. The derivative is $2x$ and (ignoring the constants) we see $x$ is also in the integral and so the substitution $u=x^2+3$ will unravel this integral!

Step 2: We must now swap the variable of integration from x to u. That means interchanging all the x's in the integrand (the term being integrated) for u's and also swapping (dx" to "du").

$u=x^2+3 \Rightarrow \frac{du}{dx}=2x \Rightarrow dx = \frac{1}{2x} du$

Then,

$\int 3x(x^2+3)^4 \ dx = \int 3x \cdot u^4 \cdot \frac{1}{2x} du = \int \frac{3}{2}u^4\ du$.

The substitution has made this integral is easy to solve!

$\int \frac{3}{2}u^4\ du= \frac{3}{10}u^5 + C$

Finally we can substitute back to get the answer in terms of x:

$\int 3x(x^2+3)^4 \ dx = \frac{3}{10}(x^2+3)^5+C$