Question

Evaluate the indefinite integral of:

\sqrt{x^2 + 2x + 10}
------------------
x +1

dx

I would like a hint please.

Answer 1

I tried doing complete the square, then trig substitution, but then I end up getting:

1
----
(cosx)^2(sinx)

Answer 2

\[\int\limits_{}^{}\frac{\sqrt{x^2+2x+10}}{x+1} dx\]
correct?

Answer 3

what is wrong with \(u=x^2+2x\)

Answer 4

i think that would result to du = (x+1) dx but x+1 is in the denom so i think that cant be used can it?

Answer 5

\[\int\limits_{}^{}\frac{\sqrt{x^2+2x+1+10-1}}{x+1}dx=\int\limits_{}^{}\frac{\sqrt{(x+1)^2+3^2}}{x+1}dx\]

\[\tan(\theta)=\frac{x+1}{3} => 3 \tan(\theta)=x+1 => 3 \sec^2(\theta) d \theta=1 dx\]
So we have
\[\int\limits_{}^{}\frac{\sqrt{(3\tan(\theta))^2+3^2}}{3 \tan(\theta)} 3 \sec^2(\theta) d \theta\]
\[\int\limits_{}^{} \frac{\sqrt{3^2} \sqrt{\tan^2(\theta)+1}}{\tan(\theta)} \sec^2( \theta) d \theta \]
\[3 \int\limits_{}^{}\frac{\sec^3(\theta)}{\tan(\theta)} d \theta \]

Answer 6

\[3\int\limits_{}^{} \frac{1}{\cos^3(\theta)} \cdot \frac{\cos(\theta)}{\sin(\theta)} d \theta \]
\[3 \int\limits_{}^{}\frac{1}{\cos^2(\theta) } \cdot \frac{1}{\sin(\theta)} d \theta=3\int\limits_{}^{}\sec^2(\theta) \csc(\theta) d \theta\]
Lets try integration by parts:
\[\int\limits_{}^{}\sec^2(\theta) \csc(\theta) d \theta=\tan(\theta) \csc(\theta)-\int\limits_{}^{}\tan(\theta) \cdot (-\cot(\theta) \csc(\theta)) d \theta\]
\[\int\limits_{}^{}\sec^2(\theta) \csc(\theta) d \theta=\tan(\theta) \csc(\theta)+\int\limits_{}^{}\csc(\theta) d \theta\]
:)
=>

\[3 \int\limits_{}^{}\sec^2(\theta) \csc(\theta) d \theta=3 \tan(\theta) \csc(\theta)+3 \int\limits_{}^{}\csc(\theta) d \theta\]

Answer 7

Do you remember how to integrate csc(theta) with respect to theta?

Answer 8

\(\large \mathtt{\text{ that is some heavy}}\) \(\large \mathtt{\LaTeX ing}\)