Integration methods for rational fractions with square roots

Let be a real number.

Integrals containing

1. Fraction with a root at the denominator




2. Fraction with and a root in the denominator




3. Fraction with and a root in the denominator




4. Simple root

Integrals containing

1. Fraction with a root at the denominator

1. Setting down


2. Setting down



Both expressions having a common member, they are both equal up to a constant and we do obtain as a bonus an explicit definition of the function with :




2. Fraction with and a root in the denominator




3. Fraction with and a root in the denominator



4. Simple root

Integrals containing

1. Fraction with a root at the denominator

1. Setting down
2. Setting down

Both expressions having a common member, they are both equal up to a constant and we do obtain as a bonus an explicit definition of the function with :




2. Fraction with and a root in the denominator

3. Fraction with and a root in the denominator



4. Simple root

Integration methods and antiderivatives recap table

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Demonstrations

We seen that with derivatives of trigonometric functions we do have this:

We will study three cases based on these integrals: integrals containing , or .

Let be a real number.

Integrals containing

1. Fraction with a root at the denominator

We have a sign generator placed in the integrand, let's study its value in the study interval .

However, we know from the derivative of a reciprocal function that:

So in our case,

But we know the derivative of the function:

So, by combining the two previous expressions we obtain:

By replacing our initial variable in this sign generator, we have:

In the study interval, this ratio is always worth .

Now we integrate only:

So,




And finally,




2. Fraction with and a root in the denominator

We set down a new variable: .

But,

So,

Let us decompose this integrand in simple elements.

Then, it exists such as:




By performing , we determine :

Now by performing , we determine :

So, the the integral can now be rewritten as:




And as a result,




3. Fraction with and a root in the denominator

We set down : .

Then we have,

As well as above, this sign generator always worth :

Therefore, we only have to integrate:

We are in the case of a standard antiderivative.

We replace it by its value.

We use again :



And finally we do obtain,



4. Simple root

As well as above, this sign generator always worth :




And now we integrate this:

To integrate this trig power, we will use the trigonometric duplication formulas to linearize it.

So,

We use another trigonometric duplication formula to linearize it.

As previously, replacing by its value:

But, we already seen several times that:

By injecting it, we have now:




And as a result,

Integrals containing

1. Fraction with a root at the denominator

1. Setting down

By setting down , we do have:

We again have a sign generator placed in the integrand, let's study its value in the study interval .

The same way as:

The equivalent for hyperbolic functions is:

By replacing our initial variable in this sign generator, we have:

In any case, this ration is always worth .




So, we now integrate:




And as a result,



2. Setting down

Moreover, by setting down :

We again have a sign generator placed in the integrand, let's study its value in the study interval .

Using again the derivative of a reciprocal function, we do have:

But we know the derivative of the function:

By combining the two previous expressions, we obtain:

By replacing our initial variable in this sign generator, we have:

In any case, this ration is always worth .

So, we only integrate:

However, we do have below in the page, in the trigonometric antiderivatives that:

Using again the derivative of a reciprocal function, we do have:

But we know the derivative of the function:

By combining the two previous expressions, we obtain:

By then injecting our result into the initial expression we have:

The constant being absorbed by the main integration constant, we finally obtain that:



Both expressions and having a common member, they are equal up to a constant.

Let us determine this constant by taking a value of , for example .

So, we find that:




We then have as a bonus an explicit definition of the function with :




2. Fraction with and a root in the denominator

As previously, we set down: .

Or,

Now we integrate the following:

But, we already solved this integral above:



And finally,




3. Fraction with and a root in the denominator

We set down the new variable: .

Which gives us,

We saw above that this sign generator was always worth :




Consequently, removing it from the integrand:




We set another variable down: . We do have:

We are in the case of a standard antiderivative:

So in our case:

Now going back the variable changes in their order of assignment.

But, we saw above that:

So,

So, replacing in the previous expression:




And as a result,



4. Simple root

We can set down .




We saw above that this sign generator always worth :

Consequently, removing it from the integrand:

On peut alors faire an integration by parts :

Now, with the previous expression , we had:

Then, replacing it:

And as we know the antiderivative of , we can replace it.

Let us finally replace by its value:

But, we saw above that:

So,




The constant will be absorbed by the main constant integration and:

Integrals containing

1. Fraction with a root at the denominator



1. Setting down

By setting down , we do have:




The function being always positive, so do the one, and :




So, we now integrate only :




And finally,



2. Setting down

Moreover, by setting down , we do have:

Let us calculate the value of this sign generator placed under the integrand.

Using again the derivatives of reciprocal functions, we do have:

But, we know the derivative of the function:

By combining the two previous expressions, we obtain:

In the study interval, this ratio is worth :




We now integrate:

Let us calculate the value of before to take care of the sign:

Using again .

We can now inject the last expression into our previous expression:

The constant being absorbed by the main integration constant, we finally obtain that:

So the final integral is worth:

We then have to manage the two cases:




1. For the negative part:

is negative and the sign generator too:

The constant will be also absorbed by the main integration constant.




2. For the positive part:

is positive and the sign generator too:




3. Conclusion

In any case , the integral is worth:






Both expressions and having a common member, they are equal up to a constant in their common interval.

Let us determine this constant by taking a value of , for example .

Then, we find that:




We then have as a bonus an explicit definition of the function with :




2. Fraction with and a root in the denominator

As before, we set down: .

But,

We now integrate:

We then recognize a standard antiderivative.




And as a result,




3. Fraction with and a root in the denominator

We set down a new variable: .

And consequently,

We already calculated it above and this sign generator always worth :




So we now integrate:

We then in a case of a standard antiderivative:

Now, replacing it we obtain:

Using again the derivatives of reciprocal functions, we do have:

But we know the derivative of the function:

En combinant les deux expressions précédentes, on obtient :

So, replacing it:




As a result we obtain,



4. Simple root

We can set down the variable: :

We saw above that this sign generator was worth :

And we now integrate:




Let us firstly calculate the value of the integral before to take care of the sign:

These two integrals have already been calculated:

The first on this page:

And the second one previously:

So,

Moreover, we have also seen that:

The replacing it we do have now:

We can now factorize by the big factor containing :




The constant being absorbed by the main integration constant, we finally obtain by taking in consideration the sign generator left aside:

We then have to cases to manage now:





1. For the negative part:

is negative and the sign generator too, so:

The constant will be also absorbed by the main integration constant.




2. For the positive part:

is positive and the sign generator too, so:




3. Conclusion

In any case, , this integral is worth:

Integration methods and antiderivatives recap table