I = I_1 I_2 I_1= 4∫ t^2dt=4.t^3/3+C_1=4/3x^3/4+C_1 Now, I_2 = 4∫t^2/1+t^3dt Again, put 1+t^3 = Z ⇒ 3t^2dt = dz ⇒ t^2dt = 1/3dz = 4/3∫1/zdz =4/3log|Z|+C_2=4/ ...

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Byju's Answer

Standard XII

Mathematics

Integration by Substitution

∫x1 / 2/1+x3 ...

Question

$\int \frac{x^{1 / 2}}{1 + x^{3 / 4}} d x$

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Solution

$I = I_{1} - I_{2} I_{1} = 4 \int t^{2} d t = 4. \frac{t^{3}}{3} + C_{1} = \frac{4}{3} x^{3 / 4} + C_{1} N o w, I_{2} = 4 \int \frac{t^{2}}{1 + t^{3}} d t A g a i n, p u t 1 + t^{3} = Z \Rightarrow 3 t^{2} d t = d z \Rightarrow t^{2} d t = \frac{1}{3} d z = \frac{4}{3} \int \frac{1}{z} d z = \frac{4}{3} l o g | Z | + C_{2} = \frac{4}{3} l o g | (1 + t^{3}) | - C_{2} = \frac{4}{3} x^{3 / 4} - l o g | (1 + x^{3 / 4}) + C [∴ C = C_{1} - C_{2}]$

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Similar questions

\int \frac{x^{\frac{1}{2}}}{1 + x^{\frac{3}{4}}} d x =

Q. If

I = \int (\frac{1}{x^{1 / 2} - x^{1 / 4}} + \frac{log (1 + x^{1 / 4})}{x^{1 / 2} + x^{1 / 4}}) d x

, then I equals

\int \frac{(x^{3} + 8) (x - 1)}{x^{2} - 2 x + 4} d x

\int \frac{(x^{3} + 3) (x - 2)}{(x^{2} - 4 x + 4)} d x

Q. Solve the following:

\frac{x - 1}{x^{3 / 4} + x^{1 / 2}} \cdot \frac{x^{1 / 2} + x^{1 / 4}}{x^{1 / 2} + 1} \cdot x^{1 / 4} + 1

for

x = 16

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∫x1/21+x3/4dx

I=I1−I2I1=4∫t2dt=4.t33+C1=43x3/4+C1Now,I2=4∫t21+t3dtAgain,put1+t3=Z⇒3t2dt=dz⇒t2dt=13dz=43∫1zdz=43log|Z|+C2=43log|(1+t3)|−C2=43x3/4−log|(1+x3/4)+C [∴C=C1−C2]

$\int \frac{x^{1 / 2}}{1 + x^{3 / 4}} d x$