Question

a) integrate (3x+1)√4x^2 + 12x + 5 dx
b) integrate x^2+x+5 / (x^2+4)(x+1)

Answer 1

a) Ans:

$$\text{Let} \ I=\int ( 3x+1)\sqrt{4x^2+12x+5} \ dx$$

Let $3x+1\equiv \alpha(\text{derivative of }4x^2+12x+5)+\beta$

$\implies \ 3x+1=\alpha(8x+12)+\beta$

$\implies \ 3x+1=8\alpha x+(12\alpha+\beta)$

On equating the coefficients of constant term and variable x both side ,we get

$8\alpha=3 \ and \ 12\alpha+\beta=1$

$\alpha=\frac{3}{8} \ and \ \beta=1-12\alpha$

$=1-12×\frac{3}{8}$

$=\frac{-7}{2}$

$\therefore 3x+1=\frac{3}{8}(8x+12)-\frac{7}{2}$

$\therefore I=\frac{3}{8}\intop (8x+12)\sqrt{4x^2+12x+5} \ -\frac{7}{2}\int\sqrt{4x^2+12x+5}$

$=I_1-I_2 \ (says)$

Where, $I_1=\frac{3}{8}\int(8x+12)\sqrt{4x^2+12x+5} \ and \ I_2=\frac{7}{2}\int\sqrt{4x^2+12x+5}$

Let $4x^2+12x+5=t \ \implies\frac{dt}{dx}=8x+12$

$\implies dt=(8x+12)dx$ .

$$\therefore I_1=\frac{3}{8}\int\sqrt{t} dt \ =\frac{3}{8} × \frac{t^{1+\frac{1}{2}}}{1+\frac{1}{2}}+c_1$$

$=\frac{3}{8}×\frac{2}{3}×t^{\frac{3}{2}}+c_1 \ where \ c_1$ is a integration constant.

$=\frac{1}{4} {(4x^2+12x+5)}^{\frac{3}{2}} +c_1$

Now ,$I_2=\frac{7}{2}\int\sqrt{4x^2+12x+5}=\frac{7}{2}\int\sqrt{{(2x)}^2+2.2x.3+3^2-2^2}$

$=\frac{7}{2}\int\sqrt{{(2x+3)}^2-2^2}$

Put $2x+3=v\implies 2=\frac{dv}{dt} \implies dv=2dt$

$\therefore I_2=\frac{7}{2} × 2 \int\sqrt{v^2-2^2}$

Which is ,of the form $\int\sqrt{x^2-a^2}dx$ ,therefore

$I_2=7 (\frac{v\sqrt{v^2-2^2}}{2} -\frac{2^2}{2}log(v+\sqrt{v^2-2^2}))+c_2$

Where $c_2$ is an integration constant.

$=7[\frac{(2x+3)\sqrt{4x^2+12x+5}}{2}-2log(2x+3+\sqrt{4x^2+12x+5} \ ]+c_2$

Therefore,

$I=\frac{1}{4}×{(4x^2+12x+5)}^\frac{3}{2}-\frac{7}{2}(2x+3)\sqrt{4x^2+12x+5}+2log(2x+3+\sqrt{4x^2+12x+5})+k$

$\text{where}, k=c_1-c_2$ is a integration constant.

b)Ans:

$$\int\frac{x^2+x+5}{(x^2+4)(x+1)}=\int\frac{(x^2+4)+(x+1)}{(x^2+4)(x+1)}$$

$$=\int\frac{x^2+4}{(x^2+4)(x+1)}+\int\frac{x+1}{(x^2+1)(x+1)}$$

$$=\int\frac{1}{x+1}+\int\frac{1}{x^2+1}$$

$=log(x+1)+tan^{-1}x+c$

Where c is an integration constant.