How do I perform u-substitution on this problem?

Question

I am having trouble with this problem: $$\int {\frac{3x + 5}{5x^2 - 4x - 1}} dx$$

I can't seem to find a u where the du exists in the numerator so that it will cancel.

If I choose: $$u = 5x^2 - 4x - 1$$

Then: $$du = 10x - 4 dx$$

I'm fundamentally not clear on how to rewrite my problem so that I can perform u-substitution here.

Answer 1

A possible approach when confronted with integrals of rational functions (with a complicated polynomial in the numerator), is to "split" the rational function into two fractions for which we know the integral.

In this case, we can factor the numerator into: $$ 5x^2-4x-1=(5x+1)(x-1) $$ The idea now is to try to find $A$ and $B$ such that: $$ \frac{3x+5}{5x^2-4x-1}=\frac{A}{5x+1}+\frac{B}{x-1} $$ Because given these $A$ and $B$ we know that: $$ \int \frac{3x+5}{5x^2-4x-1}\;dx=\int \frac{A}{5x+1}+\frac{B}{x-1} \;dx=\frac{A}{5}\log(5x+1)+B\log(x-1) $$

When we add the fractions on the right we find that: $$ 3x+5=A(x-1)+B(5x+1)=(A+5B)x+(B-A) $$ From which it follows that: $$ A+5B=3 \qquad\text{and}\qquad B-A=5 $$ We can solve this to find: $$ A=-\frac{11}{3},\qquad B=\frac{4}{3} $$ Therefore the solution is: $$ \int \frac{3x+5}{5x^2-4x-1}\;dx=-\frac{11}{15}\log(5x+1)+\frac{4}{3}\log(x-1) $$

Answer 2

Notice, one should let $$M\frac{d}{dx}(5x^2-4x-1)+N=3x+5$$ $$M(10x-4)+N=3x+5$$ by comparing the corresponding coefficients on both the sides, $M=\frac{3}{10}$ & $M=\frac{31}{5}$, hence

$$\int \frac{3x+5}{5x^2-4x-1}\ dx$$ $$=\int \frac{\frac{3}{10}(10x-4)+\frac{31}{5}}{5x^2-4x-1}\ dx$$ $$=\frac{3}{10}\int \frac{(10x-4)dx}{5x^2-4x-1}+\frac{31}{5}\int \frac{1}{5x^2-4x-1}\ dx$$$$=\frac{3}{10}\int \frac{(10x-4)dx}{5x^2-4x-1}+\frac{31}{5}\int \frac{1}{5\left(\left(x-\frac{2}{5}\right)^2-\frac{9}{25}\right)}\ dx$$

$$=\frac{3}{10}\int \frac{(10x-4)dx}{5x^2-4x-1}+\frac{31}{25}\int \frac{1}{\left(x-\frac{2}{5}\right)^2-\frac{9}{25}}\ dx$$ Now, for the first integral, substitute $u=5x^2-4x-1\implies (10x-4) dx=du$ & for the second integral substitute $t=x-\frac{2}{5}\implies dx=dt$, $$=\frac{3}{10}\int \frac{du}{u}+\frac{31}{25}\int \frac{1}{t^2-\left(\frac{3}{5}\right)^2}\ dt$$ $$=\frac{3}{10}\ln|u|+\frac{31}{25}\cdot \frac{1}{2\cdot \frac{3}{5}}\ln\left|\frac{t-\frac{3}{5}}{t+\frac{3}{5}}\right|+C$$ $$=\frac{3}{10}\ln|5x^2-4x-1|+\frac{31}{30}\ln\left|\frac{5x-5}{5x+1}\right|+C$$