Let I= 1 x + [4] x dxLet x=t^ 4 On differentiating both sides, we get dx=4t^ 3 dt I= 4t^ 3 t^ 4 + [4] t^ 4 dt = 4t^3 t^ 2 +t dt =4 t^ 2 t+1 dt 4 (t-1)(t+1)+1 t+1 dt =4 [(t-1)+1 t+1 ]dt =4 [ t^2 2 -t+ (t+1) ]+C =2x-4 [4] x +4 ( [4] x+1 )+C Hence, 1 x + [4] x dx=2x-4 [4] x +4 ( [4] x+1 )+C

1 x + [4] x dx

Download App

Question

$\int\frac{1}{\sqrt{\text{x}}+\sqrt[4]{\text{x}}}\text{dx}$

✓

Answer

Let $\text{I}=\int\frac{1}{\sqrt{\text{x}}+\sqrt[4]{\text{x}}}\text{dx}$Let $\text{x}=\text{t}^{4}$
On differentiating both sides, we get $\text{dx}=4\text{t}^{3}\text{dt}$ $\therefore\ \text{I}=\int\frac{4\text{t}^{3}}{\sqrt{\text{t}^{4}}+\sqrt[4]{\text{t}^{4}}}\text{dt}$ $=\int\frac{4\text{t}^3}{\text{t}^{2}+\text{t}}\text{dt}$ $=4\int\frac{\text{t}^{2}}{\text{t}+1}\text{dt}$ $4\int\frac{(\text{t}-1)(\text{t}+1)+1}{\text{t}+1}\text{dt}$ $=4\int\Big[(\text{t}-1)+\frac{1}{\text{t}+1}\Big]\text{dt}$ $=4\Big[\frac{\text{t}^2}{2}-\text{t}+\log(\text{t}+1)\big]+\text{C}$ $=2\sqrt{x}-4\sqrt[4]{\text{x}}+4\log\big(\sqrt[4]{\text{x}+1}\big)+\text{C}$ Hence, $\int\frac{1}{\sqrt{\text{x}}+\sqrt[4]{\text{x}}}\text{dx}=2\sqrt{x}-4\sqrt[4]{\text{x}}+4\log\big(\sqrt[4]{\text{x}+1}\big)+\text{C}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "5 Marks Questions" from Indefinite Integrals→Indefinite Integrals — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

The probability distribution of a discrete random variable X is given as under:

$\text{X}$	$1$	$2$	$4$	$2\text{A}$	$3\text{A}$	$5\text{A}$
$\text{P}(\text{X})$	$\frac{1}{2}$	$\frac{1}{5}$	$\frac{3}{25}$	$\frac{1}{10}$	$\frac{1}{25}$	$\frac{1}{25}$