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nadya68 [22]
3 years ago
11

Find the integral using substitution or a formula.

Mathematics
1 answer:
Nadusha1986 [10]3 years ago
8 0
\rm \int \dfrac{x^2+7}{x^2+2x+5}~dx

Derivative of the denominator:
\rm (x^2+2x+5)'=2x+2

Hmm our numerator is 2x+7. Ok this let's us know that a simple u-substitution is NOT going to work. But let's apply some clever Algebra to the numerator splitting it up into two separate fractions. Split the +7 into +2 and +5.

\rm \int \dfrac{x^2+2+5}{x^2+2x+5}~dx

and then split the fraction,

\rm \int \dfrac{x^2+2}{x^2+2x+5}~dx+\int\dfrac{5}{x^2+2x+5}~dx

Based on our previous test, we know that a simple substitution will work for the first integral: \rm \quad u=x^2+2x+5\qquad\to\qquad du=2x+2~dx

So the first integral changes,

\rm \int \dfrac{1}{u}~du+\int\dfrac{5}{x^2+2x+5}~dx

integrating to a log,

\rm ln|x^2+2x+5|+\int\dfrac{5}{x^2+2x+5}~dx

Other one is a little tricky. We'll need to complete the square on the denominator. After that it will look very similar to our arctangent integral so perhaps we can just match it up to the identity.

\rm x^2+2x+5=(x^2+2x+1)+4=(x+1)^2+2^2

So we have this going on,

\rm ln|x^2+2x+5|+\int\dfrac{5}{(x+1)^2+2^2}~dx

Let's factor the 5 out of the intergral,
and the 4 from the denominator,

\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\frac{(x+1)^2}{2^2}+1}~dx

Bringing all that stuff together as a single square,

\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(\dfrac{x+1}{2}\right)^2+1}~dx

Making the substitution: \rm \quad u=\dfrac{x+1}{2}\qquad\to\qquad 2du=dx

giving us,

\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(u\right)^2+1}~2du

simplying a lil bit,

\rm ln|x^2+2x+5|+\frac52\int\dfrac{1}{u^2+1}~du

and hopefully from this point you recognize your arctangent integral,

\rm ln|x^2+2x+5|+\frac52arctan(u)

undo your substitution as a final step,
and include a constant of integration,

\rm ln|x^2+2x+5|+\frac52arctan\left(\frac{x+1}{2}\right)+c

Hope that helps!
Lemme know if any steps were too confusing.

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accoridng to byu idaho entorllemtn statistic there are 12000 femailes student here on camputes during any given semster of those
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In two or three sentences explain how you would solve for the real solutions of the following equation: please help asap!!
g100num [7]

the real solutions for the equation x^{3}=-64 are -

x=4,2+-2\sqrt{3i}

Step-by-step explanation:

   x^{3} = - 64

   x^{3} +64  = 0

   We can write 64 as  4^{3}

  x^{3} + 4^{3} = 0

  using the identity  ( x^{3}+y^{3} = (x+y)(x^{2} -xy+y^{2} ) )

we get,

  = (x+4) (x^{2} -x*4+4^{2} )

  = (x+4)(x^{2} -4x+16)    ....................(1)

 solving the quadratic equation  ,

   x^{2} -4x+16 =0

solutions of this quadratic equation can be obtained by

   x=-b +- \sqrt{b^{2}-4ac } /2a

let use y for factors

x=-(-4x)+-\sqrt{(-4x^{2} )-4*x^{2} *16}  / 2*x^{2}

x=4x+-\sqrt{16x^{2} -64x^{2} } /2x^{2}

x=4+-\sqrt{16-64}/2

x=4+-4\sqrt{3i} /2

<u />x=2+-2\sqrt{3i}    ..................(2)

from the equation 1 we have,

x-4=0

which gives solution x=4

and from equation 2 we got  x=2+-2\sqrt{3i}

so the real solutions for the equation x^{3}=-64 are -

x=4,2+-2\sqrt{3i}

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Answer:

55

Step-by-step explanation:

6m+90=96m

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