I = ∫3x 1/√(x^2+9)dx I = ∫3x/√(x^2+9)dx ∫1/√(x^2+9)dx I = I_1 I_2 Now, I_1=∫3x/√(x^2+9) Put x^2+9=t^2⇒2xdx=2tdt⇒ xdx=tdt ∴ I_1=3∫t/tdt =3∫ dt=3t+C_1=3√(x^ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Inequalities of Integrals

∫3 x-1/√x2+9 ...

Question

$\int \frac{3 x - 1}{\sqrt{x^{2} + 9}} d x =$

Open in App

Solution

$I = \int \frac{3 x - 1}{\sqrt{x^{2} + 9}} d x I = \int \frac{3 x}{\sqrt{x^{2} + 9}} d x - \int \frac{1}{\sqrt{x^{2} + 9}} d x I = I_{1} - I_{2} N o w, I_{1} = \int \frac{3 x}{\sqrt{x^{2} + 9}} P u t x^{2} + 9 = t^{2} \Rightarrow 2 x d x = 2 t d t \Rightarrow x d x = t d t ∴ I_{1} = 3 \int \frac{t}{t} d t = 3 \int d t = 3 t + C_{1} = 3 \sqrt{x^{2} + 9} + C_{1} a n d I_{2} = \int \frac{1}{\sqrt{x^{2} + 9}} d x = \int \frac{1}{\sqrt{x^{2} + (3)^{2}}} d x = l o g | x + \sqrt{x^{2} + 9} | + C_{2} ∴ I = 3 \sqrt{x^{2} + 9} + C_{1} - l o g | x + \sqrt{x^{2} + 9 |} - C_{2} = 3 \sqrt{x^{2} + 9} - l o g | x + \sqrt{x^{2} + 9 |} + C, w h e r e C = C_{1} - C_{2}$

Suggest Corrections

∫3x−1√x2+9dx=

I=∫3x−1√x2+9dxI=∫3x√x2+9dx−∫1√x2+9dxI=I1−I2Now, I1=∫3x√x2+9Put x2+9=t2⇒2xdx=2tdt⇒ xdx=tdt∴ I1=3∫ttdt=3∫dt=3t+C1=3√x2+9+C1and I2=∫1√x2+9dx=∫1√x2+(3)2dx=log|x+√x2+9|+C2∴ I=3√x2+9+C1−log|x+√x2+9|−C2=3√x2+9−log|x+√x2+9|+C, where C=C1−C2

$\int \frac{3 x - 1}{\sqrt{x^{2} + 9}} d x =$