Download App

Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
$\int\frac{2-3\text{x}}{\sqrt{1+3\text{x}}}\text{dx}$
✓

Answer

Let I $=\int\frac{2-3​​\text{x}}{\sqrt{1+3\text{x}}}\times\text{dx}.$ Then,
$\text{I}=\int\frac{2-3\text{x}-1+1}{\sqrt{1+3\text{x}}}\text{dx}$
$=\int\frac{-3\text{x}-1+3}{\sqrt{1+3\text{x}}}\text{dx}$
$=\int-\frac{(3\text{x}+1)}{\sqrt{1+3\text{x}}}\text{dx}+3\int\frac{1}{\sqrt{1+3\text{x}}}\text{dx}$
$=-1\int\frac{1+3\text{x}}{\sqrt{1+3\text{x}}}\text{dx}+3\int\frac{1}{\sqrt{1+3\text{x}}}\text{dx}$
$=-1\int(1+3\text{x})^\frac{1}{2}\text{dx}+3\int(1+3\text{x})^\frac{-1}{2}\text{dx}$
$=-1\times\frac{(1+3\text{x})^\frac{3}{2}}{\frac{3}{2}\times3}+3\times\frac{(1+3\text{x})^\frac{1}{2}}{\frac{1}{2}\times3}+\text{C}$
$=-\frac{2}{9}\times(1+3\text{x})^\frac{3}{2}+2(1+3\text{x})^\frac{1}{2}+\text{C}$
$=2(1+3\text{x})^\frac{1}{2}\Big[-\frac{1}{9}(1+3\text{x})^1+1\Big]+\text{C}$
$=2(1+3\text{x})^\frac{1}{2}\Big[\frac{-1-3\text{x}+9}{9}\Big]+\text{C}$
$=2(1+3\text{x})^\frac{1}{2}\Big[\frac{8-3\text{x}}{9}\Big]+\text{C}$
$=\frac{2}{9}\sqrt{1+3\text{x}}(8-3\text{x})+\text{C}$
$\therefore\text{I}=\frac{2}{9}(8-9\text{x})\sqrt{1+3\text{x}}+\text{C}$

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

Solve the following differential equation
$\sqrt{1-\text{x}^4}\text{dy}=\text{x dx}$
→
Find the area bounded by the lines y = 4x + 5, y 5 - x and 4y = x + 5.
→
Evaluate the following integrals:
$\int^\limits{(\pi)^\frac{2}{3}}_{0}\sqrt{\text{x}}\cos^2\text{x}^{\frac{3}{2}}\text{ dx}$
→
Consider $\text{f : R} - \left\{-\frac{4}{3}\right\} \rightarrow \text{R} - \left\{\frac{4}{3}\right\} \text{given by f(x)} = \frac{\text{4x + 3}}{\text{3x + 4}}.$ Show that f is bijective. Find the inverse of f and hence find $f ^{–1}(0)$ and x such that $f ^{–1}(x) = 2$.
→
For each of the differential equation in find the particular solution satisfying the given condition:$(\text{x}+\text{y})\ \text{dy}+(\text{x}-\text{y})\ \text{dx}=0;\ \text{y}=1\ \text{when}\ \text{x}=1$
→
Prove that:
$\begin{vmatrix}1&1+\text{p}&1+\text{p}+\text{q}\\2&3+2\text{p}&4+3\text{p}+2\text{p}\\3&6+3\text{p}&10+6\text{p}+3\text{q}\end{vmatrix}=1$
→
Show that the points whose position vectors are as given below are collinear:
$2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}},\ 3\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$ and $\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}}$
→
Find the area bounded by the cutve $y = 4 - x^2$​​​​​​​ and the line $y = 0, y = 3.$
→
Differentiate $\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)$ with respect to $\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big),$ if 0 < x < 1.
→
AB is a diameter of a circle and C is any point on the circle. Show that the area of $\triangle\text{ABC}$ is maximum, when it is isosceles.
→