# Thread: Intergal Question

1. ## Intergal Question

$\int \dfrac{7}{t^{3} + 5} dt$

$\int 7(t^{3} + 5}^{-1} dt$

$u = (t^{3} + 5)$

$du = 3t^{2}$

$\dfrac{7}{3t^{2}}du = 7 dt$

$\dfrac{7}{3t^{2}} \int u^{-1}$

$\dfrac{7}{3t^{2}} \dfrac{u^{0}}{0} + C$

$\dfrac{7}{3t^{2}} \ln|t^{3} + 5| + C$ Right?

2. ## Re: Intergal Question

No firstly the integral of u^(-1) is log u. secondly you are treating t^2 as a constant which is not the case. In its present form i can't think of way, I hope you have copied the question correctly.

3. ## Re: Intergal Question

Question is copied correctly

4. ## Re: Intergal Question

This is not pretty, but here's what Wolfram gets. Sign in and you can see the step-by-step solution.

5. ## Re: Intergal Question

A way through it is to let $t^{3}+5=t^{3}+a^{3}$ where $a$ is the cube root of 5.
Factorize and split into two partial fractions.
One integrates to a log and the other (after a routine bit of algebra) to a log and an inverse tangent.

6. ## Re: Intergal Question

Originally Posted by Jason76
$u = (t^{3} + 5)$
The point behind the substitution is to rewrite the integrand in terms of u only. So
$u = t^3 + 5 \implies t = (u - 5)^{1/3}$

and
$du = 3t^2 ~ dt = 3 \left ( (u - 5)^{1/3}\right ) ^2 \cdot dt$

which gives
$dt = \frac{du}{3 (u - 5)^{2/3}}$

Now you have a value for both t and dt to put into your integrand.

This substitution doesn't help you do the problem, I just wanted to run you through how to set it up properly.

-Dan

7. ## Re: Intergal Question

Try $t^3 = 5\tan^2\theta$. Then $t = \sqrt[3]{5}\tan^{2/3}\theta}$ and $dt = \dfrac{2\sqrt[3]{5}}{3}\sec^{7/3}\theta\sin^{1/3}\theta d\theta$

\begin{align*}\int \dfrac{7}{t^3+5}dt & = \dfrac{14\sqrt[3]{5}}{3}\int \dfrac{\sec^{7/3}\theta\sin^{1/3}\theta d\theta}{5(\tan^2\theta + 1)} \\ & = \dfrac{14\sqrt[3]{5}}{15}\int \tan^{1/3}\theta d\theta\end{align*}

That's an alternate possible place to begin.

8. ## Re: Intergal Question

This is the method I outlined a couple of days ago.

Let the integral equal $I$ and let $t^{3}+5 \equiv t^{3}+a^{3},$ (to avoid various fractional powers of 5 complicating the calculation).

Then $I=\int \frac{7}{t^{3}+a^{3}} dt =7 \int \frac{1}{(t+a)(t^{2}-at+a^{2})} dt$

Split the integrand into two (partial) fractions.

Let $\frac{1}{(t+a)(t^{2}-at+a^{2})} \equiv \frac{A}{(t+a)}+\frac{Bt+C}{(t^{2}-at+a^{2})},$

then $1 \equiv A(t^{2}-at+a^{2})+(t+a)(Bt+C).$

Substituting $t=-a$ and equating coefficients of $t^{2}$ and constants across the identity leads to

$A=1/3a^{2}, \quad B=-1/3a^{2} \text{ and } C=2/3a.$

Substituting in and tidying up a little produces

$I = \frac{7}{3a^{2}}\int \frac{1}{(t+a)}- \frac{t-2a}{(t^{2}-at+a^{2})} dt$

Rewrite the top line of the second fraction as $\frac{1}{2}(2t-4a) = \frac{1}{2}(2t-a-3a)$ and split this fraction into two, then

$I = \frac{7}{3a^{2}}\int \frac{1}{(t+a)}- \frac{1}{2}\frac{2t-a}{(t^{2}-at+a^{2})}+\frac{3a}{2}\frac{1}{(t^{2}-at+a^{2})} dt$

Complete the square in the denominator of the third fraction and introduce the factor $a\sqrt{3}/2$ top and bottom to arrive at

$I = \frac{7}{3a^{2}}\int \frac{1}{(t+a)}- \frac{1}{2}\frac{2t-a}{(t^{2}-at+a^{2})}+\sqrt{3}\frac{a\sqrt{3}/2}{(t-a/2)^{2}+(a\sqrt{3}/2)^{2}}. dt$

Finally, integrate to get

$I=\frac{7}{3a^{2}}\left( \ln(t+a)-\frac{1}{2}\ln(t^{2}-at+a^{2}) + \sqrt{3}\arctan\left[\frac{2(t-a/2)}{a\sqrt{3}}\right]\right)+D,$ $D$ a constant of integration (and where $a$ is the cube root of 5).